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Lecture of D.E. in NKNU by 陳仁淳
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Introduction to Ordinary Differential Equations
December 5, 2013
2
Contents
1 Differential Equations and Their Solutions 5
1.1 Classification of Differential Equations; Their Origin and Application . . . . . . . . . 5
1.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions . . . . 13
2 First-Order Equations with Exact Solutions 19
2.1 Exact Differential Equations and Integrating Factors . . . . . . . . . . . . . . . . . . 19
2.2 Separable Equations and Equations Reducible to This Form . . . . . . . . . . . . . . 27
2.3 Linear Equations and Bernoulli Equations . . . . . . . . . . . . . . . . . . . . . . . . 32
2.4 Special Integrating Factors and Transformations . . . . . . . . . . . . . . . . . . . . . 38
3 Applications of First-Order Equations 49
3.1 Orthogonal Trajectories and Oblique Trajectories . . . . . . . . . . . . . . . . . . . . 49
3.2 Problems in Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.3 Rate Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4 Explicit Methods of Solving High-Order ODE 75
4.1 Basic Theory of Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . 75
4.2 The Homogeneous Linear Equation with Constant Coefficients . . . . . . . . . . . . . 83
4.3 The Method of Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . . . . 87
4.4 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.5 The Cauchy-Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.6 Linear Differential Equations and Difference Equations . . . . . . . . . . . . . . . . . 95
4 CONTENTS
5 Applications of Second-Order Linear ODE 105
5.1 The Differential Equation of Vibrations of a Mass on a Spring . . . . . . . . . . . . . 105
5.2 Free, Undamped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.3 Free, Damped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.4 Forced Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
5.5 Resonance Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
5.6 Electric Circuit Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6 Systems of Linear Differential Equations 145
6.1 Basic Theory of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
6.2 Matrices and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns . . . . . . . 157
Chapter 1
Differential Equations and Their
Solutions
1.1 Classification of Differential Equations; Their Origin
and Application
A. Classification
1. Definition (Differential Equation)
An equation involving derivatives of one or more dependent variables with respect to one or
more independent variables is called a differential equation.
2. Example
(1.1.1) xy(dy
dx)2 = 0
(1.1.2)d4x
dt4+ 5
d2x
dt2+ 3x = sin t
(1.1.3)∂v
∂s+
∂v
∂t= 0
(1.1.4)∂2u
∂x2+
∂2u
∂y2+
∂2u
∂z2= 0
3. Definition (Ordinary Differential Equation, ODE)
A differential equation involving ordinary derivatives of one or more dependent variables with
respect to a single independent variable is called an ordinary differential equation.
4. Example
6 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
(1.1.1) xy(dy
dx)2 = 0: x is the single independent variable. y is dependent.
(1.1.2)d4x
dt4+ 5
d2x
dt2+ 3x = sin t: t is the single independent variable. x is dependent.
5. Definition (Partial Differential Equation, PDE)
A differential equation involving partial derivatives of one or more dependent variables with
respect to more than one independent variable is called a partial differential equation.
6. Example
(1.1.3)∂v
∂s+
∂v
∂t= 0: s and t are independent. v is dependent.
(1.1.4)∂2u
∂x2+
∂2u
∂y2+
∂2u
∂z2= 0: x, y and z are independent. u is dependent.
7. Definition (Order)
The order of the highest ordered derivative involved in a differential equation is called the order
of the differential equation.
8. Example
(1.1.1): xy(dy
dx)2 = 0. It is the first order.
(1.1.2):d4x
dt4+ 5
d2x
dt2+ 3x = sin t. It is the fourth order.
(1.1.3):∂v
∂s+
∂v
∂t= 0. It is the first order.
(1.1.4):∂2u
∂x2+
∂u2
∂y2+
∂u2
∂z2= 0. It is the second order.
9. Definition (Linear)
A linear ordinary differential equation of order n, in the dependent variable y and the indepen-
dent variable x, is an equation that is in, or can be expressed in, the form
a0(x)dny
dxn+ a1(x)
dn−1y
dxn−1+ · · ·+ an−1(x)
dy
dx+ an(x)y = b(x), (1.1.5)
where a0 is not identically zero.
Observe
(1) that the dependent variable y and its various derivatives occur to the first degree only,
(2) that no products of y and / or any its derivatives are present, and
(3) that no transcendental functions of y and / or its derivatives occur.
10. Example
1.1 Classification of Differential Equations; Their Origin and Application 7
(1.1.6)d2y
dx2+ 5
dy
dx+ 6y = 0
(1.1.7)d4y
dx4+ x2 d
3y
dx3+ x3 dy
dx= xex.
11. Definition (Nonlinear)
A nonlinear ordinary differential equation is an ordinary differential that is not linear.
12. Example
(1.1.8)d2y
dx2+ 5
dy
dx+ 6y2 = 0, (nonlinear: 6y2)
(1.1.9)d2y
dx2+ 5(
dy
dx)3 + 6y = 0, (nonlinear: 5(
dy
dx)3)
(1.1.10)d2y
dx2+ 5y
dy
dx+ 6y = 0. (nonlinear: 5y
dy
dx)
13. Definition (Constant Coefficients)
Linear ordinary differential equations are further classified according to the nature of the coef-
ficients of the dependent variables and their derivatives.
Example:
Equation (1.1.6) is said to be linear with constant coefficients.
Equation (1.1.7) is said to be linear with variable coefficients.
B. Origin and Application of Differential Equations
1. Applicable Areas
Differential equations occur in connection with numerous problems that are encountered in various
branches of science and engineering. We indicate a few such problems in the following list, which
could easily be extended to fill many pages.
(1) The problem of determining the motion of a projectile, rocked, satellite, or planet.
(2) The problem of determining the change or current in an electric circuit.
(3) The problem of conduction of heat in a rod or in a slab.
2. Example (Free, Undamped Motion; See also Chapter 5)
(1) Hooke’s law
The magnitude of the force needed to produce a certain elongation, provided this elongation is
not too great. In mathematical form,
|F | = ks,
8 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
where F is the magnitude of the force, s is the amount of elongation, and k is a constant of
proportionality which we shall call the spring constant.
When a mass is hung upon a spring of spring constant k and thus produces an elongation of
amount s, the force F of the mass upon the spring therefore has magnitude ks. The spring at
the same time exerts a force upon the mass called the restoring force of the spring. This force
is equal in magnitude but opposite in sign to F and hence has magnitude −ks
L
Fig. (a)
L
l
m
Fig. (b)
L
l
O
x
m
Fig. (c)
(a) natural length L
(b) mass in equilibrium position; spring has stretched length L+ l.
(c) mass distance x below equilibrium position; spring has stretched length L+ l + x.
(2) Forces Acting Upon the Mass
(a) F1, the force of gravity of magnitude mg.
F1 = mg ↓ positive (1.1.11)
(b) F2, the restoring force of the spring.
F2 = −k(x+ l) ↑ negative (1.1.12)
When x = 0, F1 + F2 = 0, thus we have mg = kl.
Replacing kl by mg in (1.1.12) we see that
F2 = −kx−mg. (1.1.13)
(c) F3, the resisting force of the medium, called the damping force. For small velocities it is
approximately proportional to the magnitude of the velocity:
|F3| = a
∣
∣
∣
∣
dx
dt
∣
∣
∣
∣
(1.1.15)
1.1 Classification of Differential Equations; Their Origin and Application 9
where a > 0 is called the damping constant. When the mass is moving downward,F3 acts
in the upward direction and so F3 < 0.
F3 = −adx
dt, (a > 0) ↑ negative (1.1.16)
(d) F4, any external impressed forces that act upon the mass.
F4 = F (t) (1.1.17)
We now apply Newton’s second law, F = ma, where F = F1 + F2 + F3 + F4. We find
md2x
dt2= mg − kx−mg − a
dx
dt+ F (t) (1.1.18)
or
md2x
dt2+ a
dx
dt+ kx = F (t) (1.1.19)
(e) It is a nonhomogeneous second-order linear differential equation with constant coefficients.
(f) a = 0: If a = 0, the motion is called undamped; otherwise (a 6= 0)it is called damped.
(g) F (t) = 0: If F (t) = 0, the motion is called free; otherwise it is called forced.
(3) Free, Undamped Motion
The different equation (1.1.19) then reduces to
md2x
dt2+ kx = 0 (1.1.20)
where m(> 0) is the mass and k(> 0) is the spring constant. Let λ2 =k
m, we rewrite (1.1.20)
in the formd2x
dt2+ λ2x = 0 (1.1.21)
Hence the general solution of (1.1.21) can be written
x = c1 sinλt + c2 cosλt (1.1.22)
where c1 and c2 are arbitrary constants.
Initial conditions:
x(0) = x0, (initial displacement) (1.1.23)
x′(0) = v0, (initial velocity) (1.1.24)
Therefore
dx/dt|t=0 = x′(0) = v0 =⇒ [c1λ cosλt− c2λ sinλt]|t=0 = v0 (1.1.25)
10 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
We have
c2 = x0 and c1 =v0λ
The solution is
x(t) =v0λsin λt+ x0 cosλt (1.1.26)
An alternative form
x(t) = c[v0/λ
csin λt+
x0
ccos λt]
where c =√
(v0/λ)2 + x02. Let φ be defined by satisfying
v0/λ
c= − sin φ and
x0
c= cosφ. So
x(t) = c cos(λt + φ) (1.1.27)
=√
(v0/λ)2 + x02 cos(
√
k
m+ φ) (1.1.28)
(4) Simple Harmonic Motion
(a) The free, undamped motion of the mass is a simple harmonic motion.
(b) The constant c is called the amplitude of the motion and gives the maximum (positive)
displacement of the mass from its equilibrium position.
(c) The motion is a periodic motion.
x = c⇔√
k
mt+ φ = ±2nπ, n ∈ N ∪ 0
Therefore
t =
√
m
k(±2nπ − φ) > 0 (1.1.29)
The time interval between two successive maxima is called the period of the motion and it
is given by2π
λ=
2π√
k/m=
√
m
k· 2π.
(d) The reciprocal of the period, which gives the number of oscillations per second, is called
natural frequence (or simply frequence) of the motion and it is given by
λ
2π=
√
k/m
2π=
√
k
m· 1
2π.
(e) The number of φ is called the phase constant (or phase angle). See Fig. 1.1.
(5) Another View Point (Constant Cancellation)
Simple Harmonic Motion ⇒ x(t) = A cos(pt + α), so x′′(t) = −p2A cos(pt + α), and then
x′′(t) + x · p2 = 0. Then we have the ODE:
d2x
dt2+ p2x = 0.
1.2 Solutions 11
0 0.5 1 1.5 2
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
π /8 2π/8 3π/8 4π/8
Period π /4
Amplitude 0.28
t = −φ/8
X
t
Figure 1.1: x(t) = 0.28 cos(8t + φ) where φ=-0.46
1.2 Solutions
A. Nature of Solutions
1. Definition (Explicit and Implicit Solution)
Consider the nth-order ordinary differential equation
F
[
x, y,dy
dx, .....,
dny
dxn
]
= 0, (1.2.1)
where F is a real function of its (n+ 2) arguments x, y, ....., dny/dxn.
(1) Let f be a real function of defined ∀ x ∈ I = [a, b], a, b ∈ R and having an nth derivative (and
hence also all lower derivatives) ∀ x ∈ I. The function f is called an explicit solution of (1.2.1)
on I if it is fulfilled the following two requirements:
F [x, f(x), f ′(x), .....f (n)(x)] (A)
is defined ∀ x ∈ I, and
F [x, f(x), f ′(x), .....f (n)(x)] = 0 (B)
∀ x ∈ I.
(2) A relation g(x, y) = 0 is called an implicit solution of(1.2.1) on I if this relation defines at least
one real function f(x) on I such that f(x) is an explicit solutions of (1.10) on I.
12 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
(3) Both explicit solutions and implicit solutions will usually be called simply solutions.
2. Example 1.2.1d2y
dx2+ y = 0
Answer:
f(x) = 2 sin x+ 3 cosx.
3. Example 1.2.2
x+ ydy
dx= 0
Answer:
(1) Explicit Solutions:
f1(x) =√25− x2, ∀x ∈ I = (−5, 5)
f2(x) = −√25− x2, ∀x ∈ I = (−5, 5)
(2) Implicit Solutions:
(a) g(x, y) = x2 + y2 + 25 = 0→ No!
2x+ 2ydy
dx= 0. so
dy
dx= −x
y.
We obtain the formal identity: x + y(−xy) = 0. Thus g = 0 formally satisfies the differential
equation. We have no assurance that g = 0 defines any function that is an explicit solution
of the ODE. Moreover, y = ±√−25− x2 /∈ R. So g = 0 is not truly an implicit solution but
merely a formal solution.
(b) h(x, y) = x2 + y2 − 25 = 0→ Yes!
B. Methods of Solution
1. Closed-form Solution: Chap. 2 & 4.
2. Approximate Solution: series methods, numerical methods and graphical methods. Chap. 6 & 8.
3. Chap. 7: Systems of linear ODE.
4. Chap. 9: The Laplace transform.
C. Family of Solutions
1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions 13
1. Consider the ODEdy
dx= 2x. (1.2.2)
The function fc defined on R by
fc(x) = x2 + c (1.2.3)
is a solution of the differential equation (1.2.2). We call the constant c in (1.2.3) an arbitrary
constant or parameter and refer to the family of functions defined by (1.2.3) as a one-parameter
family of solutions of (1.2.2). We write this one-parameter family of solutions as
y = x2 + c.
2. We must not conclude that every 1st-order ODE has a so-called one-parameter family of
solutions which contains all solutions of the differential equation. For example, the first-order
differential equation∣
∣
∣
∣
dy
dx
∣
∣
∣
∣
+ |y|+ 1 = 0
has no (real) solutions.
3. The general 1st-order ODEdy
dx= f(x, y), (1.2.4)
where f is a real function, may be interpreted geometrically as defining a slop f(x, y) at every
point (x, y) at which the function f is defined.
4. Assume that the ODE (1.2.4) has a so-called one-parameter family of solutions that can be
written in the form
y = F (x, c), (1.2.5)
where c is the arbitrary constant. (1.2.5) is represented geometrically by a so-called one-parameter
family of curves in xy-plane. These curves are also called the integral curves.
1.3 Initial-Value Problems, Boundary-Value Problems, and
Existence of Solutions
A. IVP and BVP
1. Definition (IVP and BVP)
In an ODE problem, if all of the associated supplementary conditions relate to one x value, the prob-
lem is called an initial-value problem ( or one-point boundary-value problem, IVP). If the conditions
relate to two different x values, the problem is called a two-point boundary-value problem ( or simply
14 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
a boundary-value problem, BVP).
2. Example (IVP)
d2y
dx2+ y = 0, y(1) = 3, y′(1) = −4.
3. Example (BVP)
d2y
dx2+ y = 0, y(0) = 1, y′(
π
2) = −4.
4. Definition (1st-order IVP)
Consider the 1st-order differential equation
dy
dx= f(x, y), (1.3.1)
where f is a continuous function of x and y in some domain D of the xy-plane; and let (x0, y0) be a
point of D. The initial-value problem associated with (1.3.1) is to find a solution φ of (1.3.1), defined
on some interval I containing x0, and satisfying the initial condition
φ(x0) = y0.
In the customary abbreviated notation, this IVP may be written
dy
dx= f(x, y),
y(x0) = y0.
5. Example (Separation of Variables)
dy
dx= −x
y,
y(3) = 4.
Answer.
ydy = −xdx,∫
ydy = −∫
xdx, y2 = −x2 + c.
y(3) = 4, 42 = −32 + c, c = 25.
y2 = 25− x2. Thus, y =√25− x2.
1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions 15
B. Existence of Solution
1. Example (No solution)
d2y
dx2+ y = 0,
y(0) = 1, y(π) = 5.
Answer. The auxiliary equation is
m2 + 1 = 0, m = ±i.
The corresponding part of the general solution is
c1 exp(ix) + c2 exp(−ix).
From Euler’s formula,
exp(iθ) = cos(θ) + i sin(θ),
we know the general solution of the ODE is
y = φ(x) = c1 cos(x) + c2 sin(x).
y(0) = 1
y(π) = 5⇒
c1 · 0 + c2 · 1 = 1
c1 · 0 + c2 · (−1) = 5⇒
c2 = 1
−c2 = 5⇒ No solution!
2. Example (Many solutions, no uniqueness)
dy
dx= y1/3,
y(0) = 0.
Answer. y−1/3dy = dx,3
2y2/3 = x+ c1 y = (
2
3x+ c2)
3/2.
We can find many solutions for this problem.
y = φ1(x) = 0, ∀x ∈ R.
y = φ2(x) =
(23x)3/2 if x ≥ 0;
0 if x ≤ 0.y = φ3(x) =
(23x− 2)3/2 if x ≥ 3;
0 if x ≤ 3.· · ·
This problem has infinitely many solutions. Notice that y1/3 is not differentiable ar x = 0. See Fig
1.2.
16 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
0 2 4 6 8 10 12
0
2
4
6
8
10
12
14
16
18
20
← φ2(x) ← φ
3(x)
Figure 1.2: Many solutions.
3. Theorem 1.3.1 Basic Existence and Uniqueness Theorem
Hypothesis. Consider the differential equation
dy
dx= f(x, y), (1.3.1)
where
(1) the function f is continuous in the domain D and
(2) the partial derivative ∂f/∂y is also continuous in D.
Let (x0, y0) ∈ D.
Conclusion. ∃! φ, φ is also the solution of (1.3.1), defined on some interval |x− x0| ≤ h, where
h is sufficiently small, that satisfies
φ(x0) = y0.
4. Example
dy
dx= x2 + y2,
y(1) = 3.
Answer. f(x, y) = x2 + y2 and ∂f/∂y = 2y are continuous in R2. (1, 3) ∈ R
2.
Thus, there is a unique solution φ defined on |x− 1| ≤ h such that φ(1) = 3.
1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions 17
5. Example
(1)dy
dx=
y√x, y(1) = 2.
(2)dy
dx=
y√x, y(0) = 2.
Answer. f(x, y) =y
x1/2and
∂f
∂y=
1
x1/2are continuous in R
2\x ≤ 0.(1) Theorem 1.3.1 says there is a unique solution since x0 = 1 > 0.
(2) Theorem 1.3.1 says nothing since x0 = 0.
1
ydy =
1√xdx, ln y = 2
√x+ c1, y = c · exp(2
√x).
h(x) := exp(2√x) > 0, (positive)
h′(x) = exp(2√x) · 1√
x> 0, (increasing),
h′′
(x) =
exp(2√x)√
x· √x− exp(2
√x) · 1
2√x
x,
h′′
(x) = 0 ⇒ 1 =1
2√x⇒ x =
1
4. (inflection point)
0 ∼ 1/4 1/4 ∼ր ր
concave down concave up
See Fig. 1.3. Notice that h(x) is not differentiable at x = 0 even if we define h(x) as follows,
h(x) :=
exp(2√x) if x > 0
1 if x ≤ 0.
0 0.1 0.2 0.3 0.4 0.5 0.6
0
1
2
3
4
5
6
← exp(2 sqrt(x))
Figure 1.3: The graph of e2√x.
18 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
Chapter 2
First-Order Equations with Exact
Solutions
2.1 Exact Differential Equations and Integrating Factors
A. Standard Forms of First-Order Differential Equations
1. Definition
The derivative form:dy
dx= f(x, y). (2.1.1)
The differential form: M(x, y)dx+N(x, y)dy = 0. (2.1.2)
Remark.
Newton and Leibniz each used a different notation when they published their discoveries of
calculus, thereby creating a notational divide between Britain and the European continent
that lasted for more than 50 years. The Leibniz notation dy/dx eventually prevailed because
it suggests correct formulas in a natural way, the chain rule
dy
dx=
dy
du· dudx
being a good example. The symbols ”dy” and ”dx” are called differentials. Assume that f
is differentiable at a point x, define dx to be an independent variable that can have any real
value, and define dy by the formula
dy = f ′(x) dx.
If dx 6= 0, then we can divide both sides by dx to obtain
dy
dx= f ′(x).
20 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Thus, they are called the differential and derivative forms respectively.
The symbol df is another common notation for the differential of a function y = f(x). For
example, if f(x) = sin x, then we can write df = cosx dx. All of the general rules of differenti-
ation then have corresponding differential versions:
Derivative Formula Differential Formulad
dx[c] = 0 d[c] = 0
d
dx[cf ] = c
df
dxd[cf ] = c df
d
dx[f + g] =
df
dx+
dg
dxd[f + g] = df + dg
d
dx[fg] = f
dg
dx+ g
df
dxd[f + g] = fdg + gdf
d
dx
[
f
g
]
=g dfdx− f dg
dx
g2d
[
f
g
]
=g df − f dg
g2
2. Example
(1)dy
dx=
x2 + y2
x− y(x2 + y2)dx− (x− y)dy = 0.
(2)dy
dx= −sin x+ y
x+ 3y(sin x+ y)dx+ (x+ 3y)dy = 0.
B. Exact Differential Equations
1. Definition (Total Differential)
Let F be a function of two variables such that F has continuous first partial derivatives in a domain
D. The total differential dF of the function F is defined by
dF (x, y) =∂F
∂xdx+
∂F
∂xdy
for all (x, y) ∈ D. Note that : F (x1, x2 · · ·xn), n variables
dF =∂F
∂x1dx1 + · · ·+
∂F
∂xndxn =
n∑
i=1
∂F (x1, · · · , xn)
∂xidxi.
2. Example
Find dF where F (x, y) = xy2 + 2x3y.
Answer. dF = (y2 + 6x2y)dx+ (2xy + 2x3)dy
3. Definition (Exact Differential)
The expression
M(x, y)dx+N(x, y)dy (2.1.3)
2.1 Exact Differential Equations and Integrating Factors 21
is called an exact differential in a domain D if there exists a function F of two real variables such
that this expression equals to the total differential dF (x, y), ∀(x, y) ∈ D
That is, (2.1.3) is an exact differential in D if ∃ F such that
∂F
∂x(x, y) = M(x, y) and
∂F
∂y(x, y) = N(x, y) ∀(x, y) ∈ D
If M(x, y)dx+N(x, y)dy is an exact differential, then the differential equation
M(x, y)dx+N(x, y)dy = 0
is called an exact differential equation.
4. Example
(1) y2dx+ 2xydy = 0 (2) ydx+ 2xdy = 0
Answer. (1)F = xy2 exact! (2) not exact!
5. Theorem 2.1.1
Consider
M(x, y)dx+N(x, y)dy = 0 (2.1.4)
where M and N have continuous first partial derivatives at all (x, y) ∈ D
(1) If (2.1.4) is exact in D, then
∂M(x, y)
∂y=
∂N(x, y)
∂x
(2) Conversely, if∂M
∂y=
∂N
∂x∀(x, y) ∈ D, then (2.1.4) is exact.
Proof.
Part 1. ”Exact” ⇒ ∂M
∂y=
∂N
∂x
If (2.1.4) is exact in D, then ∃F such that
∂F
∂x(x, y) = M(x, y) and
∂F
∂y(x, y) = N(x, y).
Therefore∂M
∂y=
∂2F
∂y∂x=
∂2F
∂x∂y=
∂N
∂x.
22 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Part 2. ”Exact” ⇐ ∂M
∂y=
∂N
∂x
Let us assume that∂F
∂x= M . Then
F (x, y) =
∫
M(x, y)dx+ φ(y)
∂F
∂y=
∂
∂y
∫
M(x, y)dx+dφ
dy(y)
In order to prove the differential equation is exact, we must solve φ(y) to satisfy
N(x, y) =∂
∂y
∫
M(x, y)dx+dφ
dy(y).
Hencedφ
dy= N(x, y)− ∂
∂y
∫
M(x, y)dx.
(1) We prove that N(x, y)− ∂∂y
∫
M(x, y)dx must be independent of x.
Claim : ∂∂x[N(x, y)− ∂
∂y
∫
M(x, y)dx] = 0
LHS =∂
∂xN(x, y)− ∂2
∂x∂y
∫
M(x, y)dx
=∂
∂xN − ∂2
∂x∂y[F (x, y)− φ(y)]
=∂
∂xN − ∂2
∂y∂x[F (x, y)− φ(y)]
=∂
∂xN − ∂2
∂y∂x
∫
M(x, y)dx
=∂
∂xN − ∂
∂yM
= 0
Therefore, dφdy
= N − ∂∂y
∫
M(x, y)dx is independent of x.
(2) We solve φ(y).
φ(y) =
∫ [
N(x, y)− ∂
∂y
∫
M(x, y) dx
]
dy.
Therefore,
F (x, y) =
∫
M(x, y)dx+
∫[
N(x, y)− ∂
∂y
∫
M(x, y) dx
]
dy.
6. Example
(1) y2dx+ 2xydy = 0 (2) ydx+ 2xdy = 0
2.1 Exact Differential Equations and Integrating Factors 23
Answer. (1)∂M
∂y= 2y =
∂N
∂xExact!
F =∫
y2dx+∫
[2xy − ∂∂y
∫
y2dx]dy
= xy2 +∫
[2xy − 2xy]dy
= xy2 + 0
dF = y2dx+ 2xydy. (Solutions : xy2 = c.)
(2)∂M
∂y= 1 6= 2 =
∂N
∂xNot exact!
7. Example
(2x sin y + y3ex)dx+ (x2 cos y + 3y2ex)dy = 0
Answer.∂M
∂y= 2x cos y + 3y2ex =
∂N
∂xExact!
F =
∫
(2x sin y + y3ex)dx+
∫[
x2 cos y + 3y2ex − ∂
∂y
∫
(2x sin y + y3ex)dx
]
dy
= x2 sin y + y3ex + 0
(Solutions :x2 sin y + y3ex = c.)
8. Theorem 2.1.2 (The Solution of Exact Differential Equations)
Suppose the differential equationM(x, y)dx+N(x, y)dy = 0 satisfies the differentiability requirements
of Theorem 2.1.1 and is exact in D. Then a one-parameter family of solutions of this ODE is given
by
F (x, y) = c
where F is the function such that
∂F
∂x= M and
∂F
∂y= N ∀(x, y) ∈ D
and c is an arbitrary constant.
Proof. Mdx+Ndy = 0 is exact. ∃ F such that∂F
∂x= M and
∂F
∂y= N
∂F
∂xdx+
∂F
∂ydy = 0 or simply dF (x, y) = 0. The relation
F (x, y) = c
is obviously a solution of this, where c is an arbitrary constant.
9. Example
(3x2 + 4xy)dx+ (2x2 + 2y)dy = 0
24 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Answer.∂M
∂y= 4x =
∂N
∂x
F =
∫
(3x2 + 4xy)dx+ φ(y) = x3 + 2x2y + φ(y)
∂F
∂y= 2x2 +
dφ
dy= N = 2x2 + y
dφ
dy= 2y, φ(y) = y2 + c
We may write the solution as x3 + 2x2y + y2 = c.
Method of Grouping
(3x2 + 4xy)dx+ (2x2 + 2y)dy = 0
3x2dx+ (4xydx+ 2x2dy) + 2ydy = 0
d(x3) + d(2x2y) + d(y2) = d(c)
x3 + 2x2y + y2 = c.
Notice that it requires a good ”working knowledge”!
Example: (x3 + xy2 − y)dx+ (x2y + y3 + x)dy = 0.
Answer. x(x2 + y2)dx− ydx+ y(x2 + y2)dy + xdy = 0.
(x2 + y2)(xdx+ ydy) + (xdy − ydx) = 0
xdx+ ydy +xdy − ydx
x2 + y2= 0
d(x2
2) + d(
y2
2) +
d(y/x)
1 + (y/x)2= 0
x2
2+
y2
2+ tan−1 y
x= c,
where we use the formula
∫
1
1 + u2du = tan−1 u+ c.
2.1 Exact Differential Equations and Integrating Factors 25
The proof is
tan(tan−1 u) = u
d tan(tan−1 u)
du=
du
du= 1
sec2(tan−1 u)d tan−1 u
du= 1
(1 + u2)d tan−1 u
du= 1
d tan−1 u
du=
1
1 + u2.
10. Example
Solve the IVP: (2x cos y + 3x2y)dx+ (x3 − x2 sin y − y)dy = 0, y(0) = 2.
Answer.∂M
∂y= −2x sin y + 3x2 =
∂N
∂x.
(1) Standard method
F =
∫
Mdx+ φ(y)
= x2 cos y + x3y + φ(y)
∂F
∂y= −x2 sin y + x3 +
dφ(y)
dy= N(x, y) = x3 − x2 sin y − y
dφ
dy= −y ⇒ φ(y) = −y
2
2+ c0
F (x, y) = x2 cos y + x3y − y2
2+ c0
y(0) = 2 ⇒ 0 + 0− 4/2 + c0 = 0 ⇒ c0 = 2
x2 cos y + x3y − y2
2= −2.
(2) Method of grouping
(2x cos ydx− x2 sin ydy) + (3x2ydx+ x3dy)− ydy = 0
d(x2 cos y) + d(x3y)− d(y2/2) = d(c)
x2 cos y + x3y − y2
2= c.
Of course y(0) = 2 again yields the particular solution already obtained.
11. Definition (Integrating Factor)
If the differential equation
M(x, y)dx+N(x, y)dy = 0 (2.1.5)
26 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
is not exact in the domain, D, but the differential equation
µ(x, y)M(x, y)dx+ µ(x, y)N(x, y)dy = 0 (2.1.6)
is exact in D, then µ(x, y) is called an integrating factor of the differential equation (2.1.5).
12. Example
ydx− xdy = 0.
Answer.
(1) µ =1
x2:
y
x2dx− 1
xdy = 0
∂M
∂y=
1
x2=
∂N
∂x
(2) µ =1
y2:
1
ydx− x
y2dy = 0
∂M
∂y= − 1
y2=
∂N
∂x
Notice that µ = xy and µ = 1/(xy) are also integrating factors of the ODE. Integrating factor is not
unique.
13. Example
(3y + 4xy2)dx+ (2x+ 3x2y)dy = 0.
Answer.∂M
∂y= 3 + 8xy 6= 2 + 6xy =
∂N
∂x. Consider the integrating factor µ = x2y.
(3x2y2 + 4x3y3)dx + (2x3y + 3x4y2)dy = 0
∂[µM ]
∂y= 6x2y + 12x3y2 =
∂[µN ]
∂x
14. Remark
We referred to this resulting exact equation as ”essentially equivalent” to the original. However, the
multiplication may result in either
(1) the loss of (one or more) solutions of the original, or
(2) the gain of (one or more) functions which are solutions of the ”new” equation but not of the
original, or
(3) both of these phenomena.
The question now arises: how is an integrating factor found? We will answer this question in Sec.
2.4.
2.2 Separable Equations and Equations Reducible to This Form 27
2.2 Separable Equations and Equations Reducible to This
Form
A. Separable Equations
1. Definition (Separable equations)
An equation of the form
F (x)G(y)dx+ f(x)g(y)dy = 0
is called an equation with variables separable or simply a separable equation.
2. Example
(x− 4)y4dx− x3(y2 − 3)dy = 0.
3. Theorem 2.2.1
The separable differential equation
F (x)G(y)dx+ f(x)g(y)dy = 0
has an integrating factor
µ(x, y) =1
f(x)G(y).
Proof.
µF (x)G(y)dx+ µf(x)g(y)dy = 0
F (x)
f(x)dx+
g(y)
G(y)dy = 0
∂M
∂y=
∂[F (x)/f(x)]
∂y= 0 =
∂N
∂x=
∂[g(y)/G(y)]
∂x
4. Remark
If y0 ∈ R such that G(y0) = 0, then y = y0 is a (constant) solution of the original differential
equation; and this solution may (or may not) have been lost in the formal separation process.
5. Example
(x− 4)y4dx− x3(y2 − 3)dy = 0
28 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Answer.
(x− 4)
x3dx− (y2 − 3)
y4dy = 0
(x−2 − 4x−3)dx− (y−2 − 3y−4)dy = 0
We have the one-parameter family of solutions
−1
x+
2
x2+
1
y− 1
y3= c.
y = 0 is a solution of the original equation. We lose it in the separation process.
6. Example
x sin ydx+ (x2 + 1) cos ydy = 0
y(1) = π/2.
Answer.
x
x2 + 1dx+
cos y
sin ydy = 0
1
2
∫
1
x2 + 1d(x2 + 1) +
∫
1
sin yd(sin y) = 0
ln |x2 + 1|+ 2 ln | sin y| = c0.
More neater,
ln[
|x2 + 1| · | sin y|2]
= ln c
(x2 + 1) sin2 y = c.
Now consider the solutions of sin y = 0. These are given by
y = nπ, n ∈ Z.
Each of these constant solutions y = nπ is a member of the solution
(x2 + 1) sin2 y = c,
for c = 0. Thus none of these solutions was lost in the separation process.
y(1) = π/2 ⇒ (12 + 1) · sin2 π/2 = c. We have the solution (x2 + 1) sin2 y = 2.
B. Homogeneous Equations
1. Definition (Homogeneous Equations)
2.2 Separable Equations and Equations Reducible to This Form 29
The first-order differential equation M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous if, when
written is the derivative formdy
dx= f(x, y),
there exists a function g such that f(x, y) can be expressed in the form g(y/x).
2. Example
(x2 − 3y2)dx+ 2xydy = 0
Answer.dy
dx=
3y2 − x2
2xy=
3y
2x− x
2y=
3
2
(y
x
)
− 1
2
(
1
y/x
)
.
3. Example
(y +√
x2 + y2)dx− xdy = 0
Answer.dy
dx=
y +√
x2 + y2
x=
y
x±√
x2 + y2√x2
=y
x±√
1 +(y
x
)2
.
depending on the sign of x.
4. Definition (Homogeneous Functions)
A function F is called homogeneous of degree n if
F (tx, ty) = tnF (x, y).
Example
(1) F (x, y) = x2 + 2xy + y2, F (tx, ty) = t2F (x, y).
(2) F (x, y) = x3 + 5x2y, F (tx, ty) = t3F (x, y).
5. Theorem 2.2.2
If M and N in M(x, y)dx + N(x, y)dy = 0 are both homogeneous functions of the same degree n,
then the differential equation is a homogeneous differential equation.
Proof. Let t = 1/x. Then
M(1
x· x, 1
x· y) =
(
1
x
)n
M(x, y)
M(1,y
x) =
(
1
x
)n
M(x, y)
M(x, y) = xnM(1,y
x), N(x, y) = xnN(1,
y
x).
30 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Thus we havedy
dx= −M(x, y)
N(x, y)= −x
nM(1, y/x)
xnN(1, y/x)= −M(1, y/x)
N(1, y/x).
6. Theorem 2.2.3
If
M(x, y)dx+N(x, y)dy = 0 (2.24)
is a homogeneous differential equation, then the change of variables y = vx transforms (2.24) into a
separable equation in the variables v and x.
Proof. Since Mdx+Ndy = 0 is homogeneous, it may be written in the form
dy
dx= g(
y
x).
Let y = vx (i.e., v =y
x). We have
dy
dx= v + x
dv
dx
v + xdv
dx= g(v)
[v − g(v)] dx+ xdv = 01
v − g(v)dv +
1
xdx = 0
∫
1
v − g(v)dv +
∫
1
xdx = c,
where c is an arbitrary constant. Let F (v) denote
∫
1
v − g(v)dv. Thus
F (v) + ln |x| = c
F (y
x) + ln |x| = c.
7. Example
(x2 − 3y2)dx+ 2xydy = 0
Answer. Since M and N are homogeneous of the same degree 2, the ODE is homogeneous.
dy
dx= − x
2y+
3y
2x.
2.2 Separable Equations and Equations Reducible to This Form 31
Let y = vx. We obtain
v + xdv
dx= − 1
2v+
3v
2
xdv
dx= − 1
2v+
v
22v
v2 − 1dv =
1
xdx
ln |v2 − 1| = ln |x|+ c1
|v2 − 1| = |cx|
|y2
x2− 1| = |cx|
|y2 − x2| = |cx3|.
If y ≥ x > 0, then we have y2 − x2 = cx3.
8. Example
(y +√
x2 + y2)dx− xdy = 0, y(1) = 0.
Answer. Since M and N are homogeneous of the same degree 1, the ODE is homogeneous.
dy
dx=
y +√
x2 + y2
x.
Since the initial x value is 1, we consider only x > 0 and take x =√x2.
dy
dx=
y
x+
√
1 +(y
x
)2
.
Let y = vx and obtain
v + xdv
dx= v +
√1 + v2
xdv
dx=√1 + v2
1√v2 + 1
dv =1
xdx
∫
1√v2 + 1
dv =
∫
1
xdx+ c1
ln |v +√v2 + 1| = ln |x|+ ln |c|,
where we use the formula∫
1√u2 + 1
dv = sinh−1 u+ c = ln |u+√u2 + 1|+ c.
32 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
The proof is as follows. Let y = sinh−1 u. We have
u = sinh y =ey − e−y
22 · ey · u = (ey)2 − 1
ey =2u+
√4u2 + 4
2= u+
√u2 + 1
y = sinh−1 u = ln |u+√u2 + 1|
d sinh−1 u
du=
1 + u/√u2 + 1
u+√u2 + 1
=1√
u2 + 1.
Another viewpoint:
d(v +√v2 + 1) = (1 +
2v
2√v2 + 1
)dv =v +√v2 + 1√
v2 + 1dv
1√v2 + 1
dv =1
v +√v2 + 1
d(v +√v2 + 1).
Therefor we obtain
v +√v2 + 1 = cx
y
x+
√
y2
x2+ 1 = cx
y +√
x2 + y2 = cx2 y(1) = 0 ⇒ c = 1
y +√
x2 + y2 = x2 y =1
2(x2 − 1).
2.3 Linear Equations and Bernoulli Equations
A. Linear Equations
1. Definition (Linear ODE)
A first-order ordinary differential equation is linear in the dependent variable y and the independent
variable x if it is, or can be, written in the form
dy
dx+ P (x)y = Q(x). (2.3.1)
Example: xdy
dx+ (x+ 1)y = x3 or
dy
dx+ (1 +
1
x)y = x2
2. Theorem 2.3.1
The linear differential equationdy
dx+ P (x)y = Q(x). (2.3.1)
2.3 Linear Equations and Bernoulli Equations 33
has an integrating factor of the form
exp[
∫
P (x)dx].
A one-parameter family of solutions of this equation is
y · exp[∫
P (x)dx] =
∫
exp
(∫
P (x)dx
)
·Q(x)dx+ c;
that is,
y = exp[−∫
P (x)dx] ·[∫
exp
(∫
P (x)dx
)
·Q(x)dx+ c
]
.
Furthermore, it can be shown that this one-parameter family of solutions of the equation (2.3.1)
includes of all solutions of (2.3.1)
Proof. Let us write (2.3.1) in the form
[P (x)y −Q(x)]dx+ dy = 0 (2.3.2)
We know that∂
∂y[P (x)y −Q(x)] = P (x) and
∂
∂x[1] = 0.
Equation (2.3.2) is not exact unless P (x) = 0. Let us multiply (2.3.2) by µ(x), obtaining
[µ(x)P (x)y − µ(x)Q(x)]dx+ µ(x)dy = 0 (2.3.3)
∂
∂y[µ(x)P (x)y − µ(x)Q(x)] =
∂
∂x[N(x, y) ≡ µ(x)]
µ(x)P (x) =d
dx[µ(x)]
P (x)dx =1
µdµ
∫
P (x)dx = ln |µ| or µ = exp[
∫
P (x)dx], (2.3.4)
where it is clear that µ > 0. Multiplying (2.3.1) by (2.3.4) gives
exp[
∫
P (x)dx] · dydx
+ exp[
∫
P (x)dx] · P (x)y = exp[
∫
P (x)dx] ·Q(x)
which is preciselyd
dx[exp(
∫
P (x)dx) · y] = Q(x) · exp[∫
P (x)dx].
Integrating this we obtain
exp(
∫
P (x)dx) · y =
∫
exp
[∫
P (x)dx
]
·Q(x)dx+ c,
y = exp
[
−∫
P (x)dx
]
·[∫
exp
(∫
P (x)dx
)
·Q(x)dx+ c
]
.
34 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
3. Example
dy
dx+ (
2x+ 1
x)y = e−2x (2.3.5)
Answer. An integrating factor is
exp[
∫
p(x)dx] = exp[
∫
(2x+ 1
x)dx]
= exp(2x+ ln | x |)= exp(2x) · x.
Multiplying (2.31) by this, we obtain
xe2xdy
dx+ e2x(2x+ 1)y = x
d
dx(xe2xy) = x
xe2xy =x2
2+ c
y =1
2xe−2x + c · 1
xe−2x
4. Example
(x2 + 1)dy
dx+ 4xy = x (2.3.6)
y(2) = 1 (2.3.7)
Answer.
dy
dx+
4x
x2 + 1y =
x
x2 + 1(2.3.8)
An integrating factor is
exp[
∫
4x
x2 + 1dx] = exp[ln(x2 + 1) · 2] = (x2 + 1)2.
2.3 Linear Equations and Bernoulli Equations 35
Multiplying (2.3.8) by this, we obtain
(x2 + 1)2dy
dx+ 4x(x2 + 1)y = x(x2 + 1)
d
dx[(x2 + 1)2y] = x3 + x
(x2 + 1)2y =1
4x4 +
x2
2+ c
y(2) = 1 ⇒ c = 19
(x2 + 1)2y =1
4x4 +
x2
2+ 19
y = (1
4x4 +
x2
2+ 19)/(x2 + 1)2.
5. Example (not linear in y but linear in x)
y2dx+ (3xy − 1)dy = 0. (2.3.9)
Answer.dy
dx=
y2
1− 3xy
It is not linear in y (not exact, not separable and not homogeneous). The roles of x and y are
interchangeable.
dx
dy=
1− 3xy
y2
dx
dy+ (
3
y)x =
1
y2. (2.3.10)
It is linear in x. An integrating factor is
exp[
∫
p(y)dx] = exp(
∫
3
ydy) = exp(ln |y|3) = y3.
Multiplying (2.3.10) by this, we obtain
y3dx
dy+ 3y2x = y,
d
dy[y3x] = y
y3x =1
2y2 + c, x =
1
2y+
c
y3.
B. Bernoulli Equations
36 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
1. Definition (Bernoulli Equations)
An equation of the formdy
dx+ P (x)y = Q(x)yn (2.3.11)
is called a Bernoulli differential equation.
2. Theorem 2.3.2
Suppose n 6= 0 and n 6= 1. Then the transformation v = y1−n reduces the Bernoulli equation (2.3.11)
to a linear equation in v.
Proof. First, we multiply (2.3.11) by y−n. thereby expressing it in the form
y−n dy
dx+ P (x)y1−n = Q(x). (2.3.12)
Let v = y1−n.dv
dx= (1− n)y−n dy
dx. From (2.38) we have
1
1− n
dv
dx+ P (x) · v = Q(x).
dv
dx+ (1− n)P (x) · v = (1− n)Q(x)
P1(x) := (1− n)P (x), Q1(x) := (1− n)Q(x)
dv
dx+ P1(x) · v = Q1(x).
This is a linear equation in v.
3. Example
dy
dx+ y = xy3
Answer. y−3 dy
dx+ y−2 = x. Let v = y1−3 = y−2. We have
dv
dx= −2y−3 dy
dx. Therefore
−12
dv
dx+ v = x
dv
dx− 2v = −2x. (2.3.13)
An integrating factor is
exp[
∫
−2dx] = e−2x.
Multiplying (2.40) by e−2x, we find
e−2x dv
dx− 2e−2xv = −2xe−2x
d
dx(e−2xv) = −2xe−2x
e−2xv =
∫
−2xe−2xdx+ c
2.3 Linear Equations and Bernoulli Equations 37
u = x dv = −2e−2xdx
du = dx v = e−2x∫
−2xe−2x = xe−2x −∫
e−2x
e−2xv = (x+1
2)e−2x + c
v = x+1
2+ ce2x.
But v =1
y2. Therefore
1
y2= x+
1
2+ ce2x.
4. General Form of the Bernoulli Differential Equation
Consider the equationdf(x)
dy· dydx
+ P (x) · f(y) = Q(x) (2.3.14)
where f is a known function of y. Then
(1) The Bernoulli differential equation (2.3.11) is a special case of (2.3.14).
(2) The transformation v = f(y) reduces (2.3.14) to a linear equation in v.
Proof.
(1) Let f(y) = y1−n.
(1− n)y−n · dydx
+ P1(x) · y1−n = Q1(x)⇒dy
dx+ P (x)y = Q(x)yn
where P1(x) = (1− n)P (x) and Q1(x) = (1− n)Q(x). Therefore, (2.3.14) ⇒ (2.3.11)
(2) Let v = f(y). We have
dv
dx=
dv
dy
dy
dx=
df(y)
dy· dydx
dv
dx+ P (x) · v = Q(x)
which is linear in v.
5. Example
cos ydy
dx+
1
xsin y = 1
Answer. Let f(y) = sin y. Therefore, df(y)dy
= cos y
df(x)
dy· dydx
+1
x· f(y) = 1.
38 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Let v = sin y. We havedv
dx+
1
xv = 1 (2.3.15)
An integrating factor is
exp[
∫
1
xdx] = exp[ln |x|] = x
Multiply (2.3.15) by x ,we find
xdv
dx+ v = x
d(xv) = x
xv =
∫
xdx+ c =1
2x2 + c
But v = sin y. Therefore, x sin y = 12x2 + c ( or sin y = 1
2x+ c
x).
Check :d
dx(x sin y) =
d
dx(1
2x2 + c)
d
dx(sin y) =
d
dx(1
2x+
c
x)
sin y + cos ydy
dxx = x+ 0 cos y
dy
dx=
1
2+−cx2
=1
2− 1
x2(sin y − x
2) · x
=1
2− sin y
x+
1
2
cos ydy
dx+
1
xsin y = 1
2.4 Special Integrating Factors and Transformations
A. Finding Integrating Factors
1. Theorem 2.4.1
Consider the differential equation
M(x, y)dx+N(x, y)dy = 0 (2.4.1)
(1) If1
N(x, y)
[
∂M(x, y)
∂y− ∂N(x, y)
∂x
]
(2.4.2)
depends upon x only, then
exp
∫
1
N(x, y)
[
∂M(x, y)
∂y− ∂N(x, y)
∂x
]
dx
(2.4.3)
2.4 Special Integrating Factors and Transformations 39
is an integrating factor of Equation (2.4.1).
(2) If1
M(x, y)
[
∂N(x, y)
∂x− ∂M(x, y)
∂y
]
(2.4.4)
depends upon y only, then
exp
∫
1
M(x, y)
[
∂N(x, y)
∂x− ∂M(x, y)
∂y
]
dy
(2.4.5)
is an integrating factor of Equation (2.4.1).
Proof.
(1) An integrating factor µ depends upon x alone.
µMdx+ µNdy = 0∂
∂y[µ(x)M(x, y)] =
∂
∂x[µ(x)N(x, y)]
µ(x) · ∂M∂y
= µ(x) · ∂N∂x
+N · dµ∂x
dµ
dx= µ(x)
[
1
N(∂M
∂y− ∂N
∂x)
]
(depends upon x only)
dµ
µ=
1
N
[
∂M
∂y− ∂N
∂x
]
dx
µ(x) = exp
∫
1
N
[
∂M
∂y− ∂N
∂x
]
dx
.
(2) An integrating factor µ depends upon y alone.
µMdx+ µNdy = 0∂
∂y· [µ(y)M(x, y)] =
∂
∂x· [µ(y)N(x, y)]
dµ
dy·M + µ(y) · ∂M
∂y= µ(y) · ∂N
∂x
dµ
dy= µ(y)
[
1
M(∂N
∂x− ∂M
∂y)
]
(depends upon y only)
dµ
µ=
1
M
[
∂N
∂x− ∂M
∂y
]
dy
µ(y) = exp
∫
1
M
[
∂N
∂x− ∂M
∂y
]
dy
.
40 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
2. Example
(2x2 + y)dx+ (x2y − x)dy = 0 (2.4.6)
It is not exact, not separable, not homogeneous, not linear, and not Bernoulli.
Answer.1
N
[
∂M
∂y− ∂N
∂x
]
=1
x2y − x[1− (2xy − 1)] = −2
x.
This depends upon x only, and so
µ(x) = exp
(
−∫
2
xdx
)
= exp(−2 ln |x|) = 1
x2
Multiplying (2.4.6) by this, we obtain
(2 +y
x2)dx+ (y − 1
x)dy = 0 (2.4.7)
By the method for solving exact ODE, we have
F (x, y) =
∫
(2 +y
x2)dx+ φ(y) = 2x− y
x+ φ(y)
∂F
∂y= −1
x+ φ′(y) = y − 1
x
φ(y) =y2
2+ c.
The solution is
2x− y
x+
y2
2= c
B. Special Transformations
1. Theorem 2.4.2
Consider the equation
(a1x+ b1y + c1)dx+ (a2x+ b2y + c2)dy = 0 (2.4.8)
where a1, b1, a2, b2, c1 and c2 are constants.
Case 1. Ifa2a16= b2
b1, then the transformation
x = X + h,
y = Y + k,
2.4 Special Integrating Factors and Transformations 41
where (h, k) is the solution of the system
a1h + b1k + c1 = 0,
a2h + b2k + c2 = 0,
reduces Equation (2.4.8) to the homogeneous equation
(a1X + b1Y )dX + (a2X + b2Y )dY = 0
in the variables X and Y .
Case 2. Ifa2a1
=b2b1
= k , then the transformation
z = a1x+ b1y
reduces Equation (2.4.8) to a separable equation in x and z.
Proof.
Case 1.
(a1X + a1h+ b1Y + b1k + c1)dX + (a2x+ a2h+ b2Y + b2k + c2)dY = 0
(a1X + b1Y )dX + (a2X + b2Y )dY = 0
It is homogeneous. We can take Y = vX to obtain
(a1 + b1Y
X)dX + (a2 + b2
Y
X)dY = 0
dY
dX= −a1 + b1(
YX)
a2 + b2(YX)
v +Xdv
dX= −a1 + b1v
a2 + b2v
Then it is a separable equation.
Case 2. dz = a1dx+ b1dy
(z + c1)dx+ (kz + c2)(1
b1)(dz − a1dx) = 0
(b1z + b1c1)dx+ (kz + c2)dz − (a1kz + a1c2)dx = 0
(b1z + a1kz + b1c1 + a1c2)dx+ (kz + c2)dz = 0
dx+kz + c2
b1z + a1kz + b1c1 + a2c2dz = 0
It is separable.
2. Example
(x− 2y + 1)dx+ (4x− 3y − 6)dy = 0 (2.4.9)
42 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Answer.
h− 2k + 1 = 0
4h− 3k + (−6) = 0
We have h=3, k=2.
Let
x = X + 3
y = Y + 2This reduce (2.53) to
(X − 2Y )dX + (4X − 3Y )dY = 0
dY
dX=
1− 2( YX)
3( YX− 4)
Let Y = vX to obtain
v +Xdv
dX=
1− 2v
3v − 43v − 4
3v2 − 2v − 1dv = − 1
XdX
∫
3v − 4
3v2 − 2v − 1dv =
1
2ln |3v2 − 2v − 1| − 3
4ln
∣
∣
∣
∣
3v − 3
3v + 1
∣
∣
∣
∣
(1) The Use of Tables.
∫
1
ax2 + bx+ cdx =
2√4ac− b2
tan−1 2ax+ b√4ac− b2
if b2 − 4ac < 0
1√b2 − 4ac
ln
(
2ax+ b−√b2 − 4ac
2ax+ b+√b2 − 4ac
)
if b2 − 4ac > 0
∫
x
ax2 + bx+ cdx =
1
2aln(ax2 + bx+ c)− b
2a
∫
1
ax2 + bx+ cdx
Then we have
∫
3v − 4
3v2 − 2v − 1dv = 3
∫
v
3v2 − 2v − 1dv − 4
∫
1
3v2 − 2v − 1dv
= 3 ·[
1
6ln |3v2 − 2v − 1| − −2
6
∫
1
3v2 − 2v − 1dv
]
− 4
∫
1
3v2 − 2v − 1dv
=1
2ln |3v2 − 2v − 1| − 3 ·
∫
1
3v2 − 2v − 1dv
=1
2ln |3v2 − 2v − 1| − 3 · 1√
4 + 12ln
∣
∣
∣
∣
6x− 2− 4
6x− 2 + 4
∣
∣
∣
∣
=1
2ln |3v2 − 2v − 1| − 3
4ln
∣
∣
∣
∣
3v − 3
3v + 1
∣
∣
∣
∣
2.4 Special Integrating Factors and Transformations 43
Moreover, multiplying 4
→ ln(3v2 − 2v − 1)2 − 3
[
ln | v − 1
3v + 1|+ ln 3
]
→ ln(3v2 − 2v − 1)2 − ln
∣
∣
∣
∣
v − 1
3v + 1
∣
∣
∣
∣
3
= ln
∣
∣
∣
∣
(3v + 1)5
v − 1
∣
∣
∣
∣
since 3v2 − 2v − 1 = (3v + 1)(v − 1)
(2) Partial Fraction
3v − 4
3v2 − 2v − 1=
a
3v + 1+
b
v − 1where a =
15
4and b =
−14
∫
3v − 4
3v2 − 2v − 1dv =
∫ 154
3v + 1dv +
∫ −14
v − 1dv
=5
4
∫
1
3v + 1d(3v + 1) + (
−14)
∫
1
v − 1d(v − 1)
=5
4ln |3v + 1| − 1
4ln |v − 1|
→ multiplying 4
5 ln |3v + 1| − ln |v − 1|
= ln
∣
∣
∣
∣
(3v + 1)5
v − 1
∣
∣
∣
∣
1
2ln |3v2 − 2v − 1| − 1
4ln |3v − 3
3v + 1|3 = − ln |X|+ C2
ln |3v2 − 2v − 1|2 − ln | v − 1
3v + 1|3 = − ln |X|4 + ln(C1)
4
ln
∣
∣
∣
∣
(3v + 1)5
v − 1
∣
∣
∣
∣
= ln((C1)
4
X4)
Finally, X4 · |(3v + 1)5| = C · |v − 1| where C = (C1)4. Now replacing v by Y/X , we obtain
|3Y +X|5 = C|Y −X|
Moreover, replacing X by x− 3 and Y by y − 2, we obtain
|3(y − 2) + (x− 3)|5 = C · |y − 2− x+ 3||x+ 3y − 9|5 = C · |y − x+ 1|
44 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
3. Example
(x+ 2y + 3)dx+ (2x+ 4y − 1)dy = 0 (2.4.10)
Answer. Let z = x+ 2y. Then we can transform the Equation (2.4.10) into
(z + 3)dx+ (2z − 1)(dz − dx
2) = 0
7dx+ (2z − 1)dz = 0
7x+ z2 − z = c.
Replacing z by x+ 2y we obtain
7x+ (x+ 2y)2 − (x+ 2y) = c
x2 + 4xy + 4y2 + 6x− 2y = c
4. Special Transformation I: v = ln y
Example.
x · dydx− y ln y = x2y (y > 0) (2.4.11)
Answer. Let v = ln y.dv
dx=
1
y· dydx
. Then we can transform (2.1) into
x · y dvdx− y · v = x2y
dv
dx− 1
x· v = x.
An integrating factor is
exp
(
−∫
1
xdx
)
= exp (− ln |x|) = 1
x.
Then we have
1
x· dvdx− 1
x2· v = 1
d(1
x· v) = 1,
v
x= x+ c
v = x · (x+ c)
Replacing v by ln y we obtain
ln y = x · (x+ c), y = ex(x+c).
2.4 Special Integrating Factors and Transformations 45
5. Special Transformation II: u = x+ y, v = x/y
Example.x
y(dx+ dy) + (x+ y)(ydx− xdy) = 0 (2.4.12)
Answer. Let u = x+ y and v =x
y. We can transform (2.4.12) into
v · du+ u · ydx− xdy
y2· y2 = 0
v · du+ u · dv · 1
(v + 1)2· u2 = 0
1
u3du+
1
v(v + 1)2dv = 0
∫
1
v(v + 1)2dv = ln |v| − ln |v + 1|+ 1
v + 1
1
v(v + 1)2=
1
v+−1v + 1
+−1
(v + 1)2∫
1
v(v + 1)2dv = ln |v| − ln |v + 1|+ 1
v + 1
Then we have
−u−2 · 12+ ln |v| − ln |v + 1|+ 1
v + 1= c
Replacing u by x+ y and v by x/y, we obtain
− 1
2(x+ y)2+ ln | x/yx
y+ 1|+ 1
xy+ 1
= c
− 1
2(x+ y)2+ ln | x
x+ y|+ y
x+ y= c
6. Special Transformation III: v = xy
Example.
xdy + ydx = (x3y2 − x)dx (2.4.13)
46 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Answer. Let v = xy. dv = ydx+ xdy. We can transform (2.4.13) into
dv = (x · v2 − x)dx,1
v2 − 1dv = xdx
∫
1
v2 − 1dv = [
∫
1
v − 1− 1
v + 1dv] · 1
2=
1
2· ln |v − 1
v + 1|
Therefore,
1
2· ln |v − 1
v + 1| =
1
2x2 + c1
ln |v − 1
v + 1| = x2 + c2 (c2 = 2c1)
v − 1
v + 1= exp[x2 + c2] (c = exp[c2])
xy − 1
xy + 1= c · ex2
7. Special Transformation IV: x = r cos θ, y = r sin θ
Example.
y2(xdx+ ydy) + x(ydx− xdy) = 0 (2.4.14)
Answer. Let
x = r cos θ therefore x2 + y2 = r2
y = r sin θ y/x = tan θ x/y = cot θ.
xdx+ ydy =1
2d(x2 + y2) =
1
2d(r2)
ydx− xdy = y2d(x
y) = y2 · d(cot θ)
Then we can transform (5.4) into
y2 · (12dr2) + x · (y2 · d cot θ) = 0
1
2r · 2 · dr + r cos θ · (− csc2 θ)dθ = 0
dr + (− cos θ csc2 θ)dθ = 0
dr + (− cot θ csc θ)dθ = 0
dr + d(csc θ) = 0
2.4 Special Integrating Factors and Transformations 47
Therefore,
r + csc θ = c, r + 1sin θ
= c
r sin θ + 1 = c · sin θ, y + 1 = c · sin θr(y + 1) = c · r sin θ, r(y + 1) = c · yr2(y + 1)2 = c2y2, (x2 + y2)(y + 1)2 = c2y2.
Check:
(2x+ 2ydy
dx)(y + 1)2 + (x2 + y2) · 2(y + 1)
dy
dx= 2c2y
dy
dx(xdx+ ydy)(y + 1)2 + (x2 + y2) · (y + 1)dy = c2ydy
(xdx+ ydy)(y + 1)2 + (x2 + y2) · (y + 1)dy =1
y2· (x2 + y2)(y + 1)2ydy
Divide it by y + 1 and we have
(xdx+ ydy)(y + 1) + (x2 + y2)dy =1
y· (x2 + y2)(y + 1)dy
xy(y + 1)dx+ [y3 + y2 + y(x2 + y2)− (x2 + y2)(y + 1)]dy = 0
(xy2 + xy)dx+ (y3 − x2)dy = 0.
48 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS
Chapter 3
Applications of First-Order Equations
3.1 Orthogonal Trajectories and Oblique Trajectories
A. Orthogonal Trajectories
1. Definition (Angle of Intersection)
By the angle of intersection of two coplanar curves at a point which they have in common is meant
−1.5 −1 −0.5 0 0.5 1 1.50.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
θ
Figure 3.1: Angle of intersection
the angles between the tangents to the curves at the common point. If the angles of intersection are
50 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
right angles, the two curves are said to be orthogonal. See Fig. 3.1.
2. Definition (Orthogonal Trajectories)
Let
F (x, y, c) = 0 (3.1.1)
be a given one-parameter family of curves in the xy-plane. A curve that intersects the curve of family
(3.1.1) at right angles is called an orthogonal trajectory of the given family.
3. Example 3.1.1
Consider the family of circles
F (x, y, c) = x2 + y2 − c2 = 0 (3.1.2)
with center at the origin and radius c. Each straight line through the origin,
y = kx, (3.1.3)
is an orthogonal trajectory of (3.1.2). Conversely, each circle of (3.1.2) is an orthogonal trajectory of
(3.1.3). See Fig. 3.2.
Electric Lines Equipotential Lines
Figure 3.2: The orthogonal trajectory of (3.1.2) and (3.1.3)
4. Procedure for Finding the Orthogonal Trajectories of a Given Family of Curves
3.1 Orthogonal Trajectories and Oblique Trajectories 51
Step1. From the equation
F (x, y, c) = 0 (3.1.1)
of the given family of curves, find the differential equation
dy
dx= f(x, y). (3.1.4)
Step2. Solvedy
dx= − 1
f(x, y).
Step3. Obtain a one-parameter family
G(x, y, c) = 0 or y = F (x, c).
Notice that
(1) For certain trajectories that are vertical lines and must be determined separately.
(2) In Step1, in finding the differential equation (3.1.4), be sure to eliminate the parameter c during
the process.
5. Example 3.1.2
F (x, y, c) = x2 + y2 − c2 = 0
Answer.
Step 1. Rewrite
2x+ 2ydy
dx= 0. So
dy
dx= −x
y.
Step 2. Solvedy
dx=
y
x.
1
ydy =
1
xdx.
ln |y| = ln |x|+ k. So y = cx where c = ek.
Step 3. Obtain G(x, y, c) = y − cx = 0. See Fig. 3.2.
6. Example 3.1.3
F (x, y, c) = y − cx2 = 0 (3.1.5)
52 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
Answer.
Step1.dy
dx− 2cx = 0 (3.1.6)
Eliminating the parameter c between (3.1.5) and (3.1.6),we have
dy
dx= 2(
y
x2)x
Step2. Solvedy
dx= − x
2y2ydy = −xdx
y2 = −12x2 + k
2y2 + x2 = c2 where c2 = 2k
Step3. G(x, y, c) = x2 + 2y2 − c2 = 0. See Fig. 3.3.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Figure 3.3: The orthogonal trajectories of (3.1.5) and (3.1.6)
7. Example 3.1.4
F (x, y, c) = x2 + (y − c)2 = c2 (3.1.7)
3.1 Orthogonal Trajectories and Oblique Trajectories 53
Answer.
Step1.
2x+ 2(y − c)dy
dx= 0 ,
dy
dx=−xy − c
(3.1.8)
Eliminating the parameter c between (3.1.7) and (3.1.8), we have
dy
dx=
−xy − x2+y2
2y
=2xy
x2 − y2
Step2. Solvedy
dx= −x
2 − y2
2xy= −1− ( y
x)2
2( yx)
Let y = vx.dy
dx= v + x
dv
dx. So we have v + x
dv
dx= −1 − v2
2v, x
dv
dx= −1 + v2
2v,
2v
1 + v2dv = −1
xdx
ln |1 + v2| = − ln |x|+ c1
1 + v2 = c2 ·1
x, x2 + y2 = c2x := +2cx (c :=
c22)
(x− c)2 + y2 = c2
Step3. G(x, y, c) = (x− c)2 + y2 − c2 = 0. See Fig. 3.4.
−8 −6 −4 −2 0 2 4 6 8−6
−4
−2
0
2
4
6
Figure 3.4: The orthogonal trajectories of (3.1.7) and (3.1.8)
B. Oblique Trajectories
54 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
1. Definition (Oblique or Isogonal Trajectories)
Let
F (x, y, c) = 0 (3.1.1)
be a one-parameter family of curves. A curve that intersects the curves of the family (3.1.1) at a
constant angle α 6= π2is called an oblique trajectories of the given family.
2. Theorem
dy
dx= f(x, y) and
dy
dx=
f(x, y) + tanα
1− f(x, y) tanα
ordy
dx= f(x, y) and
dy
dx=
f(x, y)− tanα
1 + f(x, y) tanα
have a constant angle of intersection, α.
Proof. Let θ1 = tan−1[f(x, y)]. Then we have
Case 1. θ2 = tan−1[f(x, y)] + α (i.e. θ2 − θ1 = α).
tan θ2 = tan[tan−1(f(x, y)) + α]
=tan[tan−1(f(x, y))] + tanα
1− tan[tan−1(f(x, y))] tanα
=f(x, y) + tanα
1− f(x, y) tanα
Hence the slope of this oblique trajectory is given by
dy
dx=
f(x, y) + tanα
1− f(x, y) tanα.
Case 2. θ2 = tan−1[f(x, y)]− α (i.e. θ1 − θ2 = α).
tan θ2 = tan[tan−1(f(x, y))− α]
=tan[tan−1(f(x, y))]− tanα
1 + tan[tan−1(f(x, y))] tanα
=f(x, y)− tanα
1 + f(x, y) tanα
Hence the slope of this oblique trajectory is given by
dy
dx=
f(x, y)− tanα
1 + f(x, y) tanα.
3. Example 3.1.5
3.1 Orthogonal Trajectories and Oblique Trajectories 55
Find a family of oblique trajectories that intersect the family of straight line y = cx at angle 45.
Answer.
Case 1.dy
dx=
f(x, y) + tanα
1− f(x, y) tanα
dy
dx= c =
y
x
Solvedy
dx=
yx+ tanπ/4
1− yxtan π/4
=x+ y
x− y
It is homogeneous. Let y = vx.
v + xdv
dx=
1 + v
1− v,
v − 1
1 + v2dv = −1
xdx
v
1 + v2dv − 1
1 + v2dv = −1
xdx
1
2ln |1 + v2| − tan−1 v = − ln |x| − ln |c|
⇒ ln c2x2(v2 + 1)− 2 tan−1 v = 0
Replacing v by y/x, we obtain
ln[(c2)(x2 + y2)]− 2 tan−1 y
x= 0
We have
G(x, y, c) = ln[c2(x2 + y2)]− 2 tan−1 y
x= 0
Case 2.dy
dx=
f(x, y)− tanα
1 + f(x, y) tanαSolve
dy
dx=
yx− tan π/4
1 + yxtan π/4
=y − x
x+ y
It is homogeneous. Let y = vx.
v + xdv
dx=
v − 1
1 + v,
v + 1
1 + v2dv = −1
xdx
v
1 + v2dv +
1
1 + v2dv = −1
xdx
1
2ln |1 + v2|+ tan−1 v = − ln |x| − ln |c|
⇒ ln c2x2(v2 + 1) + 2 tan−1 v = 0
Replacing v by y/x, we obtain
ln[(c2)(x2 + y2)] + 2 tan−1 y
x= 0
56 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
We have
G(x, y, c) = ln[c2(x2 + y2)] + 2 tan−1 y
x= 0
4. Self-Orthogonal
A given family of curves is said to be self-orthogonal if its family or orthogonal trajectories is the
same as the given family.
5. Example 3.1.6
y2 = 2cx+ c2
Proof.
−10 −8 −6 −4 −2 0 2 4 6 8 10−10
−8
−6
−4
−2
0
2
4
6
8
10
Figure 3.5: The self-orthogonal trajectory of Example 3.1.6.
2 · dydx· y = 2c+ 0 ⇒ dy
dx· y = c
Solve c2 + 2cx− y2 = 0 and we have c =−2x±
√
4x2 + 4y2
2= −x±
√
x2 + y2
3.1 Orthogonal Trajectories and Oblique Trajectories 57
Case (1) c > 0 : c = −x+√
x2 + y2 > 0
dy
dx· y = −x+
√
x2 + y2
dy
dx=−x+
√
x2 + y2
y
We must solvedy
dx= − y
−x+√
x2 + y2
to obtain the solution curves like y2 = 2cx + c2. Let r =√
x2 + y2 ≥ 0. So x2 + y2 = r2,
2x+ 2y · dydx
= 2rdr
dxand x+ y · dy
dx= r
dr
dx. So
dy
dx=
r drdx− x
y= − y
−x+√
x2 + y2=−y−x + r
(−x+ r)
(
rdr
dx− x
)
= −y2 = −r2 + x2 = (−r + x)(r + x)
rdr
dx− x = −r − x or r = x = r cos θ
dr
dx= −1 or cos θ = 1
r = −x+ c or θ = 2nπ
r2 = x2 + y2 = x2 − 2cx+ c2 or y = 0
Thus, we have G(x, y, c) = y2 + 2cx− c2 = 0.
Check the differential equation for the original curves.
x+ c = rdr
dx, r
dr
dx= x− x+
√
x2 + y2 = r
dr
dx= 1, r = x+ c1, (
√
x2 + y2)2 = (x+ c1)2
y2 = 2c1x+ c21
G(x, y, c) = y2 − 2cx− c2 = 0.
Case (2) c < 0 : c = −x−√
x2 + y2 < 0. Similarly we have
G(x, y, c) = y2 − 2cx− c2 = 0.
58 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
See Fig. 3.5.
6. Electric Field (Advanced Engineering Mathematics)
If an electrical current is flowing in a wire along the z-axis, the resulting equipotential lines in the
xy-plane are concentric circles about the origin, and the electric line of force are the orthogonal
trajectories of these circles. Find the differential equation of the these trajectories and solve it.
Answer. See Example 3.1.1 and Example 3.1.2.
Step 1. Rewrite
2x+ 2ydy
dx= 0. So
dy
dx= −x
y.
Step 2. Solvedy
dx=
y
x.
1
ydy =
1
xdx.
ln |y| = ln |x|+ k. So y = cx where c = ek.
Step 3. Obtain G(x, y, c) = y − cx = 0 or x = 0. See Fig. 3.2.
7. Electric Field (Advanced Engineering Mathematics)
Experiments show that the electric lines of force of two opposite charges of the same strength at (-1,
0) and (1, 0) are the circles through (-1, 0) and (1,0). Show that these circles can be represented by
the equation x2 + (y − c)2 = 1 + c2. Show also that the equipotential lines (orthogonal trajectories)
are the circles (x+ c)2 + y2 = c2 − 1.
Answer. See Example 3.1.4.
Step1. Solve the equations of the circles through (-1, 0) and (1,0). We can assume that the center is
(0, c). Thus the radius can be represented by r =√
(0− 1)2 + (c− 0)2 =√1 + c2. So these circles
can be represented by the equation
x2 + (y − c)2 = 1 + c2, or x2 + y2 − 2cy = 1.
For solving the orthogonal trajectories, we consider
2x+ 2(y − c)dy
dx= 0 ,
dy
dx=−xy − c
Eliminating the parameter c, we have
dy
dx=
−xy − x2+y2−1
2y
=2xy
x2 − y2 − 1
Step2. Solve
dy
dx= −x
2 − y2 − 1
2xy
2xydy = (y2 + 1− x2)dx
3.1 Orthogonal Trajectories and Oblique Trajectories 59
Consider the transformation u = y2 and we have
xdu = (u+ 1− x2)dx xdu− udx = (1− x2)dx
xdu− udx
x2= (
1
x2− 1)dx d(
u
x) = (
1
x2− 1)dx
u
x= −1
x− x− 2c
y2
x= −1
x− x− 2c
y2 = −1− x2 − 2cx (x+ c)2 + y2 = c2 − 1
Step3. G(x, y, c) = (x+ c)2 + y2 − c2 + 1 = 0. See Fig. 3.6.
Positive Point Charge Negative Point Charge
Electric Lines
Equipotential Lines
Figure 3.6: The orthogonal trajectories.
8. Fluid Flow (Advanced Engineering Mathematics)
Another area in which orthogonal trajectories play a role is fluid flow. Here the path of a par-
ticle of fluid is called a streamline, and the orthogonal trajectories of the streamlines are called
equipotential lines. Suppose that the streamlines are xy = c. Show that the walls of the channel in
Fig. 3.6 are streamlines, so that the flow may be regarded as a flow around a corner. Find and graph
the equipotential lines.
Answer.
Step 1. xy = c. y + xdy
dx= 0.
dy
dx= −y
x.
Step 2. Solvedy
dx=
x
y. ydy = xdx.
60 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
Step 3. G(x, y, c) = x2 − y2 + c = 0.
See Fig. 3.7.
9. Temperature Field
StreamLines
Equipotential Lines
Wall
Wall
Corner
Figure 3.7: The orthogonal trajectories of the fluid flow
Curves of constant temperature T (x, y) = c in a temperature field are called isotherms. Their orthog-
onal trajectories are the curves along which heat will flow (in regions that are free of sources or sinks
of heat and are filled with a homogeneous medium). If the isotherms are given by y2+2xy− x2 = c,
what are the curves of heat flow?
Answer.
Step 1. y2 + 2xy − x2 = c. 2ydy
dx+ 2y + 2x
dy
dx− 2x = 0.
dy
dx=
x− y
x+ y.
Step 2. Solvedy
dx= −x+ y
x− y. Let u = y/x.
u+ xdu
dx= −1 + u
1− u
∫ −2u+ 2
−u2 + 2u+ 1du =
∫ −2x
dx
ln | − u2 + 2u+ 1| = ln |x−2|+ c1 −(yx)2 + 2(
y
x) + 1 =
c
x2
x2 + 2xy − y2 = c.
Step 3. G(x, y, c) = x2 + 2xy − y2 − c = 0.
10. Cauchy-Riemann Equations (Complex Analysis)
3.2 Problems in Mechanics 61
Show that if the equation u(x, y) = c represents a family of curves, a representation of its orthogonal
trajectories of the form v(x, y) = c∗ can be obtained from
∂u
∂x=
∂v
∂y,
∂u
∂y= −∂v
∂x.
These so-called Cauchy-Riemann differential equations are basic in complex analysis and will be
considered below.
Answer.
u(x, y) = c ⇒ ∂u
∂x+
∂u
∂y· dydx
= 0dy
dx= −∂u/∂x
∂u/∂y
v(x, y) = c ⇒ ∂v
∂x+
∂v
∂y· dydx
= 0dy
dx= −∂v/∂x
∂v/∂y
Now we have(
−∂u/∂x∂u/∂y
)
·(
−∂v/∂x∂v/∂y
)
=
(
− ∂v/∂y
−∂v/∂x
)
·(
−∂v/∂x∂v/∂y
)
= −1.
Thus, v(x, y) = c∗ are the orthogonal trajectories of u(x, y) = c.
11. Example
Using the Cauchy-Riemann equations, find the orthogonal trajectories of ex cos y = c.
Answer. Let u(x, y) = ex cos y = c.∂u
∂x= ex cos y =
∂v
∂y,
∂u
∂y= −ex sin y =
∂v
∂x.
∂v
∂y= ex cos y
∂v
∂x= ex sin y
v =∫
ex cos ydy + φ(x) = ex sin y + φ(x). ∂v/∂x = ex sin y + φ′(x) = ex sin y ⇒ φ′(x) = 0.
v(x, y) = ex sin y = c.
3.2 Problems in Mechanics
A. Introduction
1. Definition (Mechanics)
Mechanics may be defined as that science which describes and predicts the conditions of rest or
motion of bodies under the action of forces. It is divided into three parts: mechanics of rigid bodies,
mechanics of deformable bodies, and mechanics of fluids.
62 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
(1) The mechanics of rigid bodies is divided into statics and dynamics.
(2) As far as the resistance of the structure to failure is concerned, the mechanics of deformable
bodies are studied.
(3) The mechanics of fluids is subdivided into the study of incompressible fluids and compressible
fluids. An important subdivision of the study of incompressible fluids is hydraulics, which deals with
problems involving liquids.
2. Definition
The momentum of a body is defined to be mv.
3. Newton’s Second Law
d
dt(mv) = k · F → m
dv
dt= k · F → F =
1
kma.
The simplest systems of units are those for which k = 1.
−F = m−a .
4. Acceleration of gravity w = mg, where w is the weight of the body and g is the acceleration of gravity.
5. Instantaneous velocity v =dx
dt, where v is the instantaneous velocity.
6. Instantaneous acceleration a =dv
dt=
d2x
dt, where a is the instantaneous acceleration.
7. Three forms We can express Newton’s second law in the following three forms:
F = mdv
dt(3.2.1)
F = md2x
dt2(3.2.2)
F = mvdv
dx(3.2.3)
wheredv
dt=
dv
dx
dx
dt=
dv
dx· v.
8. Table 3.2.1
British System CGS System MKS System
force pound dyne newton
mass slug g (gram) kg (kilogram)
distance foot cm (centimeter) m (meter)
time second second second
acceleration ft/sec2 cm/sec2 m/sec2
3.2 Problems in Mechanics 63
lb., libra(e): pound(s) in weight, g: the acceleration of gravity, g = 9.8 m/sec2 = 980 cm/sec2 =
32 ft/sec2.
B. Falling Body Problems
1. Definition (Modelling)
Modelling means setting up mathematical models of physics or other systems. In this section we
consider systems that can be modelled in terms of a 1st order ODE.
2. Example (A Falling Body with Air Resistance)
A body weighting 8 lb falls from rest toward the earth from a great height. As it falls, air resistance
acts upon it, and we shall assume that this resistance (in pounds) is numerically equal to 2v, where
v is the velocity (in feet per second). Find the velocity and distance fallen at time t seconds.
Answer. (1) Modelling (Formulation)
m =w
g=
8
32=
1
4(slug). The gravity force F1 = +8(pound). The air resistance F2 = −2v (pound).
Newton’s second law becomes F1 + F2 = mdv
dt.
Earth
O
B
+
x
F1 = +8
F2 = −2v
8− 2v =1
4
dv
dtv(0) = 0. The body was initially at rest.
(2) Solution
1
8− 2vdv = 4dt, 8− 2v = c · e−8t
v(0) = 0 ⇒ c = 8, v(t) = 4(1− e−8t).
dx
dt= 4(1− e−8t), x(t) = 4(t+
1
8e−8t) + c
x(0) = 0 ⇒ c = −1/2, x(t) = 4(t+1
8e−8t)− 1
2
(3) Interpretation of Results
(a) t→∞, v → the limiting velocity 4 (ft/sec)
(b) The limiting velocity is attained in a very short time.
64 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
(c) The body will not plow through the earth. When the body reaches the earth’s surface,
the differential equation no longer apply!
3. Example (Skydiver)
A skydiver equipped with parachute and other essential equipment falls from rest toward the earth.
The total weight of the man plus the equipment is 160 lb. Before the parachute opens, the air
resistance (in pounds) is numerically equal to 1/2 v, where v is the velocity (in feet per second). The
parachute opens 5 seconds after the fall begins; after it opens, the air resistance is numerically equal
to 5/8 v2. Find the velocity of the skydiver
(A) before the parachute opens, and
(B) after the parachute opens.
Answer. (1) Modelling (Formulation)
m =w
g=
160
32= 5, F1 + F2 = m
dv
dt
(A) 160− 1
2v = 5
dv
dt, v(0) = 0
(B) 160− 5
8v2 = 5
dv
dt, v(5) = v1 where v1 is the velocity when t = 5
(2) Solution
(A)1
v − 320dv = − 1
10dt, v = 320 + ce−t/10
v(0) = 0⇒ c = −320, v(t) = 320(1− e−t/10) ∀ 0 ≤ t ≤ 5
t = 5 v1 = v(5) = 320 · (1− e−5/10).= 126
(B) 160− 5
8v2 = 5
dv
dt,
1
v2 − 256dv = −1
8dt
1
32ln
∣
∣
∣
∣
v − 16
v + 16
∣
∣
∣
∣
= − t
8+ c2, ln
∣
∣
∣
∣
v − 16
v + 16
∣
∣
∣
∣
= −4t + c1∣
∣
∣
∣
v − 16
v + 16
∣
∣
∣
∣
= ce−4t, v(t) =16(ce−4t + 1)
1− ce−4t
v(5) = 126⇒ c =110
142e20, v(t) =
16(110142
e20−4t + 1)
1− 110142
e20−4t, ∀ t ≥ 5.
(3) Interpretation of Results
(a) The limit velocity = 320 ft/sec if the parachute never opened.
(b) The limit velocity = 16 ft/sec if the parachute opens.
(c) The well-known fact is that the velocity of impact with the open parachute is a small
fraction of the impact velocity that would have occurred if the parachute had not opened.
C. Frictional Forces
1. Friction
3.2 Problems in Mechanics 65
If a body moves on a rough surface, it will encounter not only air resistance but also another resis-
tance force due to the roughness of the surface. This additional force is called friction. It is shown
in physics that the friction is given by µ ·N , where
(1) µ is a constant of proportionality called the coefficient of friction, which depends upon the rough-
ness of the given surface; and
(2) N is the normal (that is, perpendicular) force which the surface exerts on the body.
2. Example
An object weighting 48lb is released from rest at the top of a plane metal slide that is inclined 30
to the horizontal. Air resistance (in pounds) is numerically equal to one-half the velocity (in feet per
second), and the coefficient of friction is one-quarter.
(A) What is the velocity of the object 2 second after it is released?
(B) If the slide is 24 feet long, what is the velocity when the object reaches the bottom?
Answer. (1) Modelling (Formulation) m = w/g = 48/32 = 3/2.
30° 48
48 sin 30°
Figure 3.8: The diagram to explain the gravitational, the frictional and the normal forces.
(A) F1, the component on the weight parallel to the plane, has value F1 = 48 sin 30 = 24.
(B) F2, the frictional force, has value F2 = −µN = −(1/4) · (48 cos 30) = −6√3.
(C) F3, the air resistance, has numerical value F3 = −(1/2)v.We apply Newton’s second law F = ma. Here F = F1 + F2 + F3.
24− 6√3− (1/2)v = (3/2)
dv
dt, v(0) = 0.
(2) Solution
(A)3
2
dv
dt= −1
2v + 24− 6
√3,
dv
48− 12√3− v
=dt
3, v(t) = 48− 12
√3− ce−t/3.
Since v(0) = 0, we have c = 48− 12√3. Therefore
v(t) = (48− 12√3)(1− e−t/3).
66 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
t = 2⇒ v(2) = (48− 12√3)(1− e−2/3)
.= 10.2 (ft/sec).
(B) x(t) =∫
v(t) dt = (48− 12√3)(t + 3e−t/3) + c. Since x(0) = 0 we have c = −(48− 12
√3) · 3.
x(t) = (48− 12√3)(t+ 3e−t/3 − 3).
The slide is 24 feet long. The object reaches the bottom at time T determined from
24 = (48− 12√3)(T + 3e−T/3 − 3).
T.= 2.6, v
.= (48− 12
√3)(1− e−0.9)
.= 12.3 (ft/sec).
D. Velocity of Escape from the Earth
1. Definition (Escape velocity)
A body of constant mass m is projected away from the earth in a direction perpendicular to the
earth’s surface with an initial velocity v0. Assuming that there is no air resistance, but taking into
account the variation of the earth’s gravitational field with distance. The least initial velocity v0 that
is required to lift the body such that the body will not return to the earth is called the escape velocity.
2. Example
Find the minimum initial velocity of a projectile that is fired in radial direction from the earth and
is supposed to escape from the earth. Neglect the air resistance and the gravitational pull of other
celestial bodies. The radius of the earth is R, the mass of projectile is m, the gravitational force is
w(x), the origin is at the surface of the earth and the position of the projectile is x.
R
x = 0 xw(x)
Answer. (1) Modelling (Formulation)
According to Newton’s law of gravitation, the gravitational force, and therefore also the acceleration,
a, of the projectile, is proportional to1
(x+R)2.
w(x) = − k
(x+R)2
where k is a constant. w(0) = −mg ⇒ −mg = −k/R2. Therefore k = mgR2.
w(x) = − mgR2
(x+R)2.
Since there are no other forces acting in the body, the equation of motion is
ma = mdv
dt= − mgR2
(x+R)2, v(0) = v0.
3.3 Rate Problems 67
(2) Solution
Expressdv
dtin terms of
dv
dx, hence
dv
dt=
dv
dx· dxdt
= v · dvdx
. Take it into the motion equation and we
have
mvdv
dx= − mgR2
(x+R)2, vdv = − gR2
(x+R)2dx,
v2
2=
gR2
x+R+ c.
t = 0⇒ x = 0; v(0) = v0. Thus we havev202
=gR2
0 +R+ c, c =
v202− gR.
v2
2=
gR2
x+R+
v202− gR, v = ±
√2 ·
√
v202− gR+
gR2
x+R.
The plus sign must be chosen if the body is rising, and the minus sign if it is falling back to the
earth. To determine the maximum altitude ξ that the body reaches, we set v = 0 and x = ξ.
v20 − 2gR+2gR2
ξ +R= 0
2gR2 = (R + ξ)(−v20 + 2gR) = −Rv20 + 2gR2 + ξ(−v20 + 2gR)
ξ =Rv20
−v20 + 2gR.
Solving the equation for v0, we find the initial velocity required to lift the body to the altitude ξ,
namely,
v0 =
√
2gR · ξ
R + ξ.
The escape velocity ve is then found by letting ξ →∞.
ve =√
2gR.
The numerical value of ve is approximately 6.9 miles/sec = 11.1 km/sec = 39960 km/hr.
3.3 Rate Problems
In certain problems the rate at which a quantity changes is a known function of the amount present
and/or the time, and it is desired to find the quantity itself. If x denotes the amount of the quantity
present at time t, then dx/dt denotes the rate at which the quantity changes and we are at once led
to a differential equation.
A. Rate of Growth and Decay
1. Example (Radioactive nuclei decay)
The rate at which radioactive nuclei decay is proportional to the number of such nuclei that are
present in a given sample. Half of the original number of radioactive nuclei have undergone disinte-
gration in a period of 1500 years.
68 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
(A) What percentage of the original radioactive nuclei will remain after 4500 years ?
(B) In how many years will only one-tenth of the original number remain ?
Answer. (1) Modelling (Formulation)
x0: the amount of radioactive nuclei initially present.
x: the amount of radioactive nuclei present after t years.dx
dt: the rate at which the nuclei decay.
The nuclei decay at a rate proportional to the amount present.
dx
dt= −kx, k > 0 (minus sign means decreasing in x)
x(0) = x0.
Half of the original number of radioactive nuclei have undergone disintegration in 1500 years.
x(1500) =1
2x0.
(2) Solution
(A)1
xdx = −kdx x = ce−kt
x(0) = x0 ⇒ c = x0 x(t) = x0 · e−kt
x(1500) =1
2x0 ⇒
1
2x0 = x0 · e−k·1500 e−k = (
1
2)
t1500
x(t) = x0(1
2)
t1500 x(4500) = x0(
1
2)45001500 =
1
8x0.
One-eighth (i.e. 12.5 %) of the original number remain after 4500 years.
(B)1
10x0 = x0(
1
2)
t1500 , ln(
1
10) =
t
1500ln(
1
2), t =
1500 ln 10
ln 2.= 4983 years.
B. Population Growth
1. Malthusian law
Assume that the rate of change of the population is proportional to the number of individuals in it
at any time.
x0: the number of individuals in a population at time t0.
x: the number of individuals in a population at time t.dx
dt: the rate at which the population grows.
dx
dt= +rx, r > 0 (plus sign means increasing in x)
x(t0) = x0.
⇒ x(t) = x0er(t−t0)
3.3 Rate Problems 69
It can be shown that the Malthusian law turns out to be unreasonable when applied to the distance
future.
2. Logistic law
The logistic law of growth
dx
dt= rx− λx2
where r > 0 and λ > 0 are constants. The additional term −λx2 is the result of some cause that
tends to limit the ultimate growth of the population. For example, such a cause could be insufficient
living space or food supply, when the population becomes sufficiently large.
3. Example
The population x of a certain city satisfies the logistic law
dx
dt=
1
100x− 1
108x2
where time t is measured in years. Given that the population of this city is 100,000 in 1980, deter-
mine the population as a function of time t > 1980. In particular, answer the following questions:
(A) what will be the population in 2000?
(B) in what year does the 1980 population double?
(C) assuming the population model can be applied for all t > 0, how large will the population ulti-
mately be?
Answer. (1) Modelling (Formulation)
dx
dt=
1
100x− 1
108x2
x(1980) = 100, 000
(2) Solution
dx
10−2x− 10−8x2= dt, 100
(
1
x+
10−6
1− 10−6x
)
dx = dt.
Assuming that 0 < x < 106, we obtain
100[ln x− ln(1− 10−6x)] = t + c1x
1− 10−6x= c · exp( t
100)
x(t) =c · exp(t/100)
1 + 10−6c · exp(t/100)x(1980) = 100, 000 105 =
c · exp(1980/100)1 + 10−6c · exp(1980/100)
c =106
9 exp(1980/100)x(t) =
106
1 + 9 exp(19.8− t/100)
70 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
(A)
x(2000) =106
1 + 9 exp(19.8− 2000/100)=
106
1 + 9 exp(−0.2).= 119, 495
(B)
200, 000 =106
1 + 9 exp(19.8− t/100)⇒ exp(19.8− t/100) =
4
9⇒ t
.= 2061
(C)
limt→∞
x(t) = limt→∞
106
1 + 9 exp(19.8− t/100)=
106
1 + 0= 1, 000, 000
Remark:
(a) Consider the logistic equation modified from the Malthusian model.
dx
dt= h(x) · x.
We want to choose h(x) so that
• h(x) ≃ r > 0 when x is small,
• h(x) decreases as x grows larger, and
• h(x) < 0 when x is sufficiently large.
The simplest function that has these properties is
h(x) = r − λx,
where λ > 0. Then the following equation
dx
dt= r · x− λx2,
is known as the logistic equation (or the Verhulst equation).
(b) The equivalent form of the logistic equation is
dx
dt= r(1− x
k)x,
where k = r/λ. The constant r is called the intrinsic growth rate, i.e., the growth rate in the absence
of any limiting factors.
(c) Solve the ODE exactly.
dx
(1− x/k)x= r dt
(
1
x+
1/k
1− x/k
)
dx = r dt
ln |x| − ln |1− x/k| = rt+ c1
3.3 Rate Problems 71
We need the assumption: 0 < x < k (and 0 < x0 < k). So we have
x
(1− x/k)= cert
x(0) = x0 ⇒ c = x0/[1− (x0/k)]
x(t) =x0k
x0 + (k − x0)e−rt
Then we can see that limt→∞ x(t) = k. It is called the saturation level.
• y = k: It is one of the equilibrium points and an asympotically stable solution.
• y = 0: It is one of the equilibrium points and an unstable solution.
C. Mixture Problems
1. Basic concepts
A substance S is allowed to flow into a certain mixture in a container at a certain rate, and the mixture
is kept uniform by stirring. Further, in one such situation, this uniform mixture simultaneously flows
out of the container at another (generally different) rate; in another situation this may not be the
case. In either case, we seek to determine the quantity of the substance S present in the mixture at
time t.
INOUT
Tank X
Figure: the mixture brine tank
x : the amount of S present at time tdx
dt: the rate of change of x with respect to t
IN : the rate of S enters the mixture
OUT : the rate of S at which it leavesdx
dt= IN−OUT
2. Example
A tank initially contains 50 gal. of pure water. Starting at time t = 0 a brine containing 2 lb of
dissolved salt per gallon flows into the tank at the rate of 3 gal/min. The mixture is kept uniform
by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate.
(A) How much salt is in the tank at any time t > 0?
72 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
(B) How much salt is present at the end of 25 min?
(C) How much salt is present after a long time?
Answer. (1) Modelling (Formulation)
dx
dt= IN−OUT
IN = ( 2 lb/gal) · (3 gal/min) = 6 lb/min.
OUT : the rate of outflow = the rate of inflow
⇒ the tank contains 50 gal of the mixture at any time t.
⇒ the concentration of salt at time t is (x/50) lb/gal.
OUT = [ (x/50) lb/gal] · (3 gal/min) = (3x/50) lb/min.
Initially there was no salt in the tank (pure water).
⇒ x(0) = 0 .
dx
dt= IN−OUT = 6− 3x
50x(0) = 0.
(2) Solution
(A)
dx
dt= 6− 3x
50dx
100− x=
3
50dt
x(t) = 100 + c · exp(− 3t
50)
x(0) = 0 ⇒ c = −100
x(t) = 100
[
1− exp(− 3t
50)
]
.
(B) x(25) = 100[1− exp(−3 · 25/50)] .= 78 (lb).
(C) limt→∞
100[1− exp(− 3t
50)] = 100 (lb).
3. Example
A tank initially contains 50 gal. of brine in which there is dissolved 10 lb of salt. Starting at time
t = 0 a brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 5 gal/min.
The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the
tank at the slower rate 3 gal/min. How much salt is in the tank at any time t > 0?
Answer. (1) Modelling (Formulation)
dx
dt= IN−OUT
3.3 Rate Problems 73
IN = (2 lb/gal) · (5 gal/min) = 10 lb/min.
OUT : At time t = 0, the tank contains 50 gal of brine.
inflow: 5 gal/min, outflow: 3 gal/min, net gain: 5-3=2 gal/min
At the end of t minutes the amount of brine in the tank is (50 + 2t) gal.
The concentration of salt at time t is
(
x
50 + 2t
)
lb/gal.
OUT = (x
50 + 2tlb/gal ) · (3 gal/min) =
(
3x
50 + 2t
)
lb/min.
Initially there was 10 lb of salt in the tank, we have
⇒ x(0) = 10 .
dx
dt= IN−OUT = 10− 3x
50 + 2tx(0) = 10.
(2) Solution
dx
dt= 10− 3x
50 + 2tdx
dt+
3x
50 + 2t= 10
µ(t) = exp
(∫
3
50 + 2tdt
)
= (t+ 25)3/2
23/2µ(t) = 23/2 · (t+ 25)3/2 = (2t+ 50)3/2
d
dt
[
(2t+ 50)3/2x]
= 10 · (2t+ 50)3/2
x(t) = 2(2t+ 50) +c
(2t+ 50)3/2
x(0) = 10 ⇒ c = −(90)(50)3/2 = −22500√2
x(t) = 4t+ 100− 22500√2
(2t+ 50)3/2.
74 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS
Chapter 4
Explicit Methods of Solving High-Order
ODE
4.1 Basic Theory of Linear Differential Equations
A. Definition and Basic Existence Theorem
1. Definition
A linear ordinary differential equation of order n in the dependent variable y and the independent
variable x is an equation that is in, or can be expressed in, the form
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = F (x) (4.1.1)
where an(x) is not identically zero. We shall assume that a0(x), a1(x), . . . , an(x) and F (x) are con-
tinuous real functions on a ≤ x ≤ b and that an(x) 6= 0 ∀ x ∈ [a, b]. The right-hand member F (x) is
called the nonhomogeneous term. If F is identically zero, Eqn. (4.1.1) reduces to
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = 0 (4.1.2)
and is then called homogeneous. For example, n = 2, the second-order nonhomogeneous linear ODE
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = F (x) (4.1.3)
and (4.1.2) reduces to the second-order homogeneous linear ODE
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = 0 (4.1.4)
2. Examples
(1) 2nd-order:d2y
dx2+ 3x
dy
dx+ x3y = ex.
(2) 3rd-order:d3y
dx3+ x
d2y
dx2+ 3x2 dy
dx− 5y = sin x.
76 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
3. Theorem 4.1.1
Hypothesis
(1) Consider the nth-order linear ODE (4.1.1) where a0(x), a1(x), . . . , an(x) and F (x) are continuous
real functions on a ≤ x ≤ b and that an(x) 6= 0 ∀ x ∈ [a, b]
(2) Let x0 ∈ [a, b] and c0, c1, . . . , cn−1 be n arbitrary real constants.
Conclusion
∃! the solution f of (4.1.1) such that
f(x0) = c0, f′(x0) = c1, . . . , f
(n−1)(x0) = cn−1
and the solution is defined over [a, b].
4. Exampled2y
dx2+ 3x
dy
dx+ x3y = ex, y(1) = 2, y′(1) = −5.
Theorem 4.1.1 assures us that there exist a unique solution of the given problem.
5. Corollary 4.1.2
Hypothesis
Let f be a solution of nth-order homogeneous ODE
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = 0 (4.1.2)
such that
f(x0) = 0, f ′(x0) = 0, . . . , f (n−1)(x0) = 0,
where x0 ∈ [a, b].
Conclusion
Then f(x) = 0 ∀ x ∈ [a, b].
6. Example
d3y
dx3+ 2
d2y
dx2+ 4x
dy
dx+ x2y = 0, f(2) = f ′(2) = f ′′(2) = 0.
Cor. 4.1.2 assures us that f(x) = 0 ∀ x ∈ R.
B. Homogeneous Equations
1. Theorem 4.1.3 (Basic Theorem on Linear Homogeneous Differential Equations)
Hypothesis
Let f1(x), f2(x), · · · , fm(x) be any m solutions of the homogeneous ODE
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = 0 (4.1.2)
4.1 Basic Theory of Linear Differential Equations 77
Conclusion
Then c1f1(x)+c2f2(x)+ · · ·+cmfm(x) is also a solution of (4.1.2) where c0, c1, . . . , cm are m arbitrary
real constants.
2. Definition (Linear Combination)
If f1(x), f2(x), · · · , fm(x) are m given functions, and c0, c1, . . . , cm are m constants, then the expres-
sion
c1f1(x) + c2f2(x) + · · ·+ cmfm(x)
is called a linear combination of f1(x), f2(x), · · · , fm(x).
3. Theorem 4.1.3 (Restated)
Any linear combination of solutions of the homogeneous linear ODE (4.1.2) is also a solution of
(4.1.2).
4. Exampled2y
dx2+ y = 0
Answer. c1 sin x+ c2 cosx.
5. Exampled3y
dx3− 2
d2y
dx2− dy
dx+ 2y = 0
Answer. c1ex + c2e
−x + c3e2x.
6. Definition (Linear Dependence)
The n functions f1(x), f2(x), · · · , fn(x) are called linearly dependent on a ≤ x ≤ b if there exist
constants c0, c1, . . . , cn, not all zeros, such that
c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0 ∀ x ∈ [a, b].
7. Example
f1(x) = x, f2(x) = 2x
Answer. Let c1 = 2, c2 = −1. Therefore c1f1(x) + c2f2(x) = 0 ∀x ∈ R. f1(x) and f2(x) are linearly
dependent.
8. Definition (Linear Independence)
The n functions f1(x), f2(x), · · · , fn(x) are called linearly independent on a ≤ x ≤ b if they are not
linear dependent there. That is, the n functions f1(x), f2(x), · · · , fn(x) are linearly independent if
c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0⇒ c0 = c1 = · · · = cn = 0.
78 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
In other words, the only linear combination of f1(x), f2(x), · · · , fn(x) that is identically zero on [a, b]
is the trivial linear combination
0 · f1(x) + 0 · f2(x) + · · ·+ 0 · fn(x).
9. Example
f1(x) = x, f2(x) = x2, [a, b] = [0, 1]
Answer. c1x+ c2x2 = 0, ∀x ∈ [0, 1].
Let x = 1/2 ⇒ 1/2 · c1 + 1/4 · c2 = 0 Therefore c1 = c2 = 0
x = 1 ⇒ c1 + c2 = 0
f1(x) and f2(x) are linearly independent.
10. Theorem 4.1.4
The nth-order homogeneous linear ODE (4.1.2) always possesses n solutions that are linearly inde-
pendent. Further, if f1(x), f2(x), · · · , fn(x) are n linearly independent solutions of (4.1.2), then every
solution of (4.1.2) can be expressed as a linear combination
c1f1(x) + c2f2(x) + · · ·+ cnfn(x).
11. Example
d2y
dx2+ y = 0, g(x) = sin(x+ π/6), f1(x) = sin x, f2(x) = cosx.
Answer.g(x) = sin(x+ π/6) = sin x cosπ/6 + cosx sin π/6
=
√3
2sin x+
1
2cosx =
√3
2f1(x) +
1
2f2(x)
12. Definition (Fundamental Set of Solutions)
If f1(x), f2(x), · · · , fn(x) are n linearly independent solutions of (4.1.2), then the set
f1(x), f2(x), · · · , fn(x) is called a fundamental set of solutions of (4.1.2) and the function f(x)
defined by
f(x) = c1f1(x) + c2f2(x) + · · ·+ cnfn(x)
where c1, c2, . . . , cn are arbitrary constants, is called a general solution of (4.1.2).
13. Example
(1)d2y
dx2+ y = 0. (2)
d3y
dx3− 2
d2y
dx2− dy
dx+ 2y = 0
4.1 Basic Theory of Linear Differential Equations 79
Answer. (1) sin x, cos x, (2) ex, e−x, e2x.
14. Definition (Wronskian)
Let f1(x), f2(x), · · · , fn(x) be n real functions each of which has an (n−1)st derivative on [a, b]. The
determinant
W (f1, f2, . . . , fn) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
f1 f2 · · · fn
f ′1 f ′
2 · · · f ′n
· · ·f(n−1)1 f
(n−1)2 · · · f
(n−1)n
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
is called the Wronskian determinant, or simply the Wronskian of these n functions. We observe that
W (f1, . . . , fn) is itself a real function defined on [a, b]. Its values at x is denoted by W (f1, . . . , fn)(x)
or by W [f1(x), . . . , fn(x)].
15. Theorem 4.1.5
Suppose that f1, . . . , fn are n solutions of (4.1.2).
Then the n solutions f1, . . . , fn of (4.1.2) are linear independent on [a, b], i.e.,
the family of solutions
y(x) = c1f1(x) + · · ·+ cnfn(x)
with arbitrary coefficients c1, . . . , cn includes every solution of (4.1.2)
⇔ W (f1, . . . , fn) 6= 0 for some x ∈ [a, b].
16. Remark (Abel’s Theorem)
The Wronskian of n solutions f1, . . . , fn of (4.1.2) is either identically zero on [a, b] or else is never
zero on [a, b]. In particular,
if f1 and f2 are solutions of the differential equation (4.1.4) where a0(x), a1(x), and a2(x) are
continuous on an open interval I, then the Wronskian W (f1, f2)(x) is given by
W (f1, f2)(x) = c · exp[
−∫
a1(x)
a2(x)dx
]
,
where c is a certain constant that depends on f1 and f2, but not on x. Further, W (f1, f2)(x) either
is zero for all x in I (if c = 0) or else is never zero in I (if c 6= 0).
17. Example
(1)d2y
dx2+ y = 0 (2)
d3y
dx3− 2
d2y
dx2− d2y
dx2+ 2y = 0
Answer.
(1)
∣
∣
∣
∣
∣
sin x cosx
cosx − sin x
∣
∣
∣
∣
∣
= − sin2 x− cos2 x = −1 6= 0 (2)
∣
∣
∣
∣
∣
∣
∣
ex e−x e2x
ex −e−x 2e2x
ex e−x 4e2x
∣
∣
∣
∣
∣
∣
∣
= −6e2x 6= 0.
80 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
C. Reduction of Order
1. Theorem 4.1.6
Let f(x) be a nontrivial solution of the nth-order homogeneous ODE (4.1.2).
Then the transformation
y = f(x)v
reduces (4.1.2) to an (n-1)st-order homogeneous linear ODE in the dependent variable
w =dv
dx.
2. Theorem 4.1.7 (Reduction in 2nd-order ODE)
Hypothesis
Let f(x) be a nontrivial solution of the second-order homogeneous linear differential equation
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = 0 (4.1.4)
Conclusion 1.
The transformation y = f(x) · v reduces (4.1.4) to an 1st-order homogeneous linear ODE
a2(x)f(x)dw
dx+ [2a2(x)f
′(x) + a1(x)f(x)]w(x) = 0 (4.1.5)
in the dependent variable w(x), where w =dv
dx.
Conclusion 2.
The particular solution
w(x) =exp[−
∫ a1(x)a2(x)
dx]
[f(x)]2
of (4.1.5) give rise to the function v(x), where
v(x) =
∫
w(x)dx =
∫ exp[−∫ a1(x)
a2(x)dx]
[f(x)]2dx.
The function g(x) defined by g(x) = f(x)v(x) is then a solution of the 2nd-order equation (4.1.4).
Conclusion 3.
The original known solution f(x) and the ”new” solution g(x) are linearly independent solutions of
(4.1.4), and hence the general solution of (4.1.4) may be expressed as
c1f(x) + c2g(x).
3. Exampled2y
dx2+ y = 0, f(x) = sin x.
4.1 Basic Theory of Linear Differential Equations 81
Find a linearly independent solution by reduction of order.
Answer. (1) Let y = f(x) · v = sin x · v.
dy
dx= cosx · v + sin x · v
d2y
dx2= − sin x · v + 2 cosx · dv
dx+ sin x · d
2v
dx2
d2y
dx2+ y = 0 ⇒ sin x · d
2v
dx2+ 2 cosx · dv
dx= 0.
(2) Let w = dv/dx.
sin x · dwdx
+ 2 cosx · w = 01
wdw = −2 cosx
sin xdx
ln |w| = −2∫
1
sin xd(sin x) = −2 ln | sin x| w =
1
sin2 xdv
dx= w =
1
sin2 xv(x) =
∫
1
sin2 xdx =
∫
csc2 xdx = − cot x.
g(x) = f(x) · v(x) = sin x · (− cotx) = − cosx.
(3) The general solution is y(x) = c1 sin x+ c2 cos x.
4. Example
(x2 + 1)d2y
dx2− 2x
dy
dx+ 2y = 0, f(x) = x.
Find a linearly independent solution by reduction of order.
Answer. (1) Let y = f(x) · v = x · v.
dy
dx= x
dv
dx+ v
d2y
dx2= x
d2v
dx2+ 2
dv
dx
x(x2 + 1)d2v
dx2+ 2
dv
dx= 0.
(2) Let w = dv/dx.
x(x2 + 1)dw
dx+ 2w = 0
1
wdw = − 2
x(x2 + 1)dx = (−2
x+
2x
x2 + 1)dx
ln |w| = −2 ln |x|+ ln(x2 + 2) + c1 w =x2 + 1
x2· c
dv
dx= w =
x2 + 1
x2dv = (1 +
1
x2)dx
Thus v(x) = x− 1
x. Therefore g(x) = f(x) · v(x) = x · (x− 1
x) = x2 − 1. (3) The general solution is
y = c1x+ c2(x2 − 1).
82 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
D. The Nonhomogeneous Equations
1. Theorem 4.1.8
Hypothesis
(1) Let v(x) be any solution of the given (nonhomogeneous) nth-order linear ODE (4.1.1).
(2) Let u(x) be any solution of the corresponding (homogeneous) nth-order linear ODE (4.1.2).
Conclusion
Then u(x) + v(x) is also a solution of the given (nonhomogeneous) equation (4.1.1).
2. Example
y = x is a solution of the eqn.d2y
dx2+ y = x (nonhomogeneous). y = sin x is a solution of the eqn.
d2y
dx2+ y = 0 (homogeneous).
Then
y = x+ sin x is also a solution ofd2y
dx2+ y = x (nonhomogeneous).
3. Theorem 4.1.9
Hypothesis
(1) Let yp(x) be a given solution of the nth-order nonhomogeneous linear equation (4.1.1) involving
no arbitrary constants.
(2) Let
yc(x) = c1y1(x) + c2y2(x) + · · ·+ cnyn(x)
be the general solution of the corresponding homogeneous equation (4.1.2).
Conclusion
Then every solution of the nth-order nonhomogeneous equation (4.1.1) can be expressed in the form
yc + yp
That is, c1y1(x) + c2y2(x) + · · · + cnyn(x) + yp(x) for suitable choice of the n arbitrary constants
c1, c2, . . . , cn.
4. Definition
Consider the nth-order nonhomogeneous linear ODE (4.1.1) and the corresponding homogeneous
equation (4.1.2).
(1) The general solution of (4.1.2) is called the complementary function of (4.1.2). We shall denote
this by yc.
(2) Any particular solution of (4.1.1) involving no arbitrary constant is called a particular integral
of (4.1.1). We shall denote this by yp.
(3) The solution yc + yp of (4.1.1) is called the general solution of (4.1.1).
5. Example
4.2 The Homogeneous Linear Equation with Constant Coefficients 83
d2y
dx2+ y = x
Answer. yc = c1 sin x+ c2 cosx. yp = x. Then the general solution is
y = yc + yp = c1 sin x+ c2 cosx+ x.
6. Theorem 4.1.10
Hypothesis
(1) Let f1(x) be a particular integral of
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = F1(x) (4.1.6)
(2) Let f2(x) be a particular integral of
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = F2(x) (4.1.7)
Conclusion
Then k1f1(x) + k2f2(x) is a particular integral of (1) Let f1(x) be a particular integral of
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · ·+ a1(x)
dy
dx+ a0(x)y = k1F1(x) + k2F2(x) (4.1.8)
where k1 and k2 are constants.
7. Example
d2y
dx2+ y = 3x+ 5 tanx
Answer.d2y
dx2+ y = x f1(x) = x (Sec. 4.3)
d2y
dx2+ y = tan x f2(x) = − cosx ln | sec x+ tanx| (Sec. 4.4)
Thus y = 3x− 5 cosx ln | sec x+ tan x| is a particular integral.
4.2 The Homogeneous Linear Equation with Constant Co-
efficients
A. Introduction
1. Definition (Characteristic Equation)
84 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
Consider
andny
dxn+ an−1
dn−1y
dxn−1+ · · ·+ a1
dy
dx+ a0y = 0 (4.2.1)
where an, an−1, . . . , a0 are real constants. The polynomial equation in the unknown m:
anmn + an−1m
n−1 + · · ·+ a1m+ a0 = 0 (4.2.2)
is called the characteristic equation or the auxiliary equation of the equation (4.2.1).
B. Case 1. Distinct Real Roots
1. Theorem 4.2.1
Consider the equation (4.2.1) with constant coefficients. If the characteristic equation (4.2.2) has the
n distinct real roots m1, m2, . . . , mn, then the general solution of (4.2.1) is
y(x) = c1em1x + c2e
m2x + · · ·+ cnemnx
where c1, c2, . . . , cn are arbitrary constants.
2. Exampled2y
dx2− 3
dy
dx+ 2y = 0
Answer. The characteristic equation is m2 − 3m+ 2 = 0. We have (m− 1)(m− 2) = 0.
y(x) = c1ex + c2e
2x
C. Case 2. Repeated Real Roots
1. Theorem 4.2.2
(1) Consider the nth-order ODE (4.2.1) with constant coefficients. If the characteristic equation
(4.2.2) has the real root m occurring k times, then the part of the general solution of (4.2.1) corre-
sponding to this k-fold repeated root is
(c1 + c2x+ · · ·+ ckxk−1)emx
(2) If, further, the remaining roots of the characteristic equation (4.2.2) are distinct real number
mk+1, . . . , mn, then the general solution is
y = (c1 + c2x+ · · ·+ ckxk−1)emx + ck+1e
mk+1x + · · ·+ cnemnx
(3) If, however, the remaining roots of the characteristic equation (4.2.2) are also repeated, then
the parts of the general solution (4.2.1) corresponding to each of these other repeated roots are
expressions similar to that corresponding to m in part (1).
4.2 The Homogeneous Linear Equation with Constant Coefficients 85
2. Exampled2y
dx2− 6
dy
dx+ 9y = 0
Answer. The characteristic equation is m2 − 6m+ 9 = 0, (m− 3)2 = 0.
y = (c1 + c2x)e3x
3. Exampled3y
dx3− 4
d2y
dx2− 3
dy
dx+ 18y = 0
Answer. The characteristic equation is m3 − 4m2 − 3m+ 18 = 0, (m− 3)2(m+ 2) = 0.
y = (c1 + c2x)e3x + c3e
−2x
D. Case 3. Conjugate Complex Roots
1. Euler’s Formula
(1) eiθ = cos θ + i sin θ
(2) eax · eibx = e(a+ib)x
(3) eiπ + 1 = 0
2. Theorem 4.2.3
(1) Consider the nth-order ODE (4.2.1) with constant coefficients. If the characteristic equation
(4.2.2) has the conjugate roots a + bi and a − bi, neither repeated, then the part of the general
solution of (4.2.1) corresponding to these two conjugate roots may be written
eax(c1 sin bx+ c2 cos bx)
(2) If, however, a + bi and a − bi are each k-fold roots of the characteristic equation (4.2.2), then
the part of the general solution of (4.2.1) corresponding to these repeated conjugate roots may be
written
eax[(c1 + c2x+ · · ·+ ckxk−1) sin bx+ (ck+1 + ck+2x+ · · ·+ c2kx
k−1) cos bx]
Note that
k1e(a+bi)x + k2e
(a−bi)x = eax[k1ebi + k2e
−bi]
= eax[k1(cos bx+ i sin bx) + k2(cos bx− i sin bx)]
= eax[(k1 + k2) cos bx+ i(k1 − k2) sin bx]
= eax[c1 cos bx+ c2 sin bx]
86 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
where c1 = k1 + k2, c2 = i(k1 − k2).
3. Example
(1)d2y
dx2+ y = 0 (2)
d2y
dx2− 6
dy
dx+ 25y = 0
Answer.m2 + 1 = 0 m2 − 6m+ 25 = 0
m = 0± i m = 3± 4i
y = c1 sin x+ c2 cos x y = e3x(c1 sin 4x+ c2 cos 4x)
4. Example
d4y
dx4− 4
d3y
dx3+ 14
d2y
dx2− 20
dy
dx+ 25y = 0
Answer.
m4 − 4m3 + 14m2 − 20m+ 25 = 0
(m2 − 2m+ 5)2 = 0
m = 1 + 2i, 1 + 2i, 1− 2i, 1− 2i
y = ex[(c1 + c2x) sin 2x+ (c3 + c4x) cos 2x]
5. Example
d2y
dx2− 6
dy
dx+ 25y = 0 y(0) = −3, y′(0) = −1
Answer.
m2 − 6m+ 25 = 0, m = 3± 4i
y = e3x(c1 sin 4x+ c2 cos 4x)
y(0) = −3 ⇒ −3 = e0(c1 sin 0 + c2 cos 0) c2 = −3y′(0) = −1 ⇒ −1 = e0[(3c1 − 4c2) sin 0 + (4c1 + 3c2) cos 0) c1 = 2
y(x) = e3x(2 sin 4x− 3 cos 4x)
=√13e3x(
2√13
sin 4x− 3√13
cos 4x)
=√13e3x sin(4x+ φ) where sin φ = − 3√
13, cosφ =
2√13
4.3 The Method of Undetermined Coefficients 87
4.3 The Method of Undetermined Coefficients
A. Introduction; An Illustrative Example
1. Example
d2y
dx2− 2
dy
dx− 3y = 2e4x
Answer. Let yp = Ae4x, y′p = 4Ae4x, y′′p = 16Ae4x.
16Ae4x − 2(4Ae4x)− 3Ae4x = 2e4x
A =2
5, yp(x) =
2
5e4x
2. Example
d2y
dx2− 2
dy
dx− 3y = 2e3x
Answer. Let yp = Ae3x. No! It is because yc = c1e3x + c2e
−x. Now we consider that
yp = Axe3x
y′p = 3Axe3x + Ae3x
y′′p = 9Axe3x + 6Ae3x
Solve the undetermined coefficient A.
(9Axe3x + 6Ae3x)− 2(3Axe3x + Ae3x)− 3(Axe3x) = 2e3x
A =1
2, yp(x) =
1
2xe3x
B. The Method
1. Definition (UC Function)
We shall call a function a UC function if it is
either (1) a function defined by one of the following:
(i) xn, n ∈ N ∪ 0,(ii) eax, a is a constant, a 6= 0,
(iii) sin(bx+ c), b and c are constants, b 6= 0,
(iv) cos(bx+ c), b and c are constants, b 6= 0,
or, (2) a function defined as a finite product of two or more functions of these four types.
88 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
2. Definition (UC Set)
Consider a UC function f(x). The set of functions consisting of f itself and all linearly indepen-
dent UC functions of which the successive derivatives of f are either constant multiples or linear
combinations will be called the UC set of f .
3. Table 4.3.1
UC function UC set
1. xn xn, xn−1, . . . , x, 1 2. eax eax 3. sin(bx+ c) or cos(bx+ c) sin(bx+ c), cos(bx+ c) 4. xneax xneax, xn−1eax, . . . , xeax, eax 5. xn sin(bx+ c) or xn cos(bx+ c) xn sin(bx+ c), xn cos(bx+ c),
xn−1 sin(bx+ c), xn−1 cos(bx+ c),
· · ·x sin(bx+ c), x cos(bx+ c),
sin(bx+ c), cos(bx+ c) 6. eax sin(bx+ c) or eax cos(bx+ c) eax sin(bx+ c), eax cos(bx+ c) 7. xneax sin(bx+ c) or xneax cos(bx+ c) xneax sin(bx+ c), xneax cos(bx+ c),
xn−1eax sin(bx+ c), xn−1eax cos(bx+ c),
· · ·xeax sin(bx+ c), xeax cos(bx+ c),
eax sin(bx+ c), eax cos(bx+ c)
4. Outline of the UC method
Outline the UC method for finding a particular integral yp of the equation (4.1.1) with constant
coefficients
andny
dxn+ an−1
dn−1y
dxn−1+ · · ·+ a1
dy
dx+ a0y = F (x)
where F (x) is a linear combination,
F (x) = A1u1(x) + A2u2(x) + · · ·+ Amum(x)
of UC functions u1, u2, . . . , um, the Ai’s being known constants.
Step 1. UC functions: u1, u2, . . . , um → UC set: S1, S2, . . . Sm.
Step 2. Si → S ′i → S
(2)i → · · · to make sure that yc 6∈ S
(n)i .
Step 3. If S(ki)i ⊆ S
(kj)j , then omit the (identical or smaller) S
(ki)i .
Find the union of S(k1)1 , S
(k2)2 , . . . , S
(km)m .
4.3 The Method of Undetermined Coefficients 89
Step 4. Form a linear combination of the set, with unknown constant coefficients
(undetermined coefficients).
Step 5. Determine these unknown coefficients by substituting the linear combination
formed in Step 4 into the differential equation and demanding that is identically
satisfies the differential equation.
5. Example
(1)d2y
dx2− 3
dy
dx+ 2y = x2ex (2)
d2y
dx2− 2
dy
dx+ y = x2ex
Answer. The UC set of x2ex is S = x2ex, xex, ex.
(1) yc = c1ex + c2e
2x (2) yc = c1ex + c2xe
x
ex ∈ S double root ex ∈ S
S ′ = x3ex, x2ex, xex, yc 6∈ S ′ S ′′ = x4ex, x3ex, x2ex, yc 6∈ S ′′
6. Example
d2y
dx2− 2
dy
dx− 3y = 2ex − 10 sin x
Answer. yc = c1e3x + c2e
−x
Step 1. S1 = ex, S2 = sin x, cosxStep 2. yc 6∈ S1 and yc 6∈ S2
Step 3. S = S1 ∪ S2 = ex, sin x, cosxStep 4. Let yp = Aex +B sin x+ C cosx.
Step 5. y′p = Aex +B cosx− C sin x, y′′p = Aex −B sin x− C cos x
[Aex − B sin x− C cos x]− 2[Aex +B cosx− C sin x] − 3[Aex +B sin x+ C cosx]
= 2ex − 10 sin x
−4Aex + (−4B + 2C) sinx+ (−4C − 2B) cosx = 2ex − 10 sin x
A = −1/2, B = 2, C = −1
yp = −12ex + 2 sin x− cosx
y = yc + yp = c1e3x + c2e
−x − 1
2ex + 2 sin x− cos x
90 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
7. Example
d2y
dx2− 3
dy
dx+ 2y = 2x2 + ex + 2xex + 4e3x
Answer. yc = c1ex + c2e
2x
Step 1. S1 = x2, x, 1, S2 = ex, S3 = xex, ex, S4 = e3xStep 2. S2 ⊂ S3 and yc ∈ S3 → S ′
3 = x2ex, xexStep 3. S = S1 ∪ S ′
3 ∪ S4 = x2, x, 1, x2ex, xex, e3xStep 4. yp = Ax2 +Bx+ C +De3x + Ex2ex + Fxex
Step 5.
2A −3B +2C = 0
−6A +2B = 0
2A = 2
2D = 4
−2E = 2
2E −F = 1
Thus we have A = 1, B = 3, C = 7/2, D = 2, E = −1, F = −3.
y = yc + yp = c1ex + c2e
2x + x2 + 3x+ 7/2 + 2e3x − x2ex − 3xex
4.4 Variation of Parameters
A. The method
Consider the general second order linear ODE
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = F (x). (4.4.1)
The corresponding homogeneous ODE is
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = 0. (4.4.2)
The complementary function of (4.4.2) is
c1y1(x) + c2y2(x).
The method is to determine
v1(x)y1(x) + v2(x)y2(x)
4.4 Variation of Parameters 91
to be a particular solution of (4.4.1). Hence we use the name, variation of parameters.
Assume yp = v1(x)y1(x) + v2(x)y2(x). Thus y′p = v1y
′1 + v2y
′2 + v′1y1 + v′2y2.
(1) We simplify y′p by demanding that
v′1y1 + v′2y2 = 0
Thus we have y′′p = v1y′′1 + v2y
′′2 + v′1y
′1 + v′2y
′2.
(2) Impose the basic condition yp be a solution of (4.4.1).
v′1y′1 + v′2y
′2 =
F (x)
a2(x)
Since W [y1(x), y2(x)] 6= 0 for y1 and y2 are linearly independent, we can solve (1) and (2) to obtain
v′1 = − F (x)y2(x)
a2(x)W [y1(x), y2(x)]
v′2 = +F (x)y1(x)
a2(x)W [y1(x), y2(x)]
Then we obtain a particular solution
yp(x) = v1(x)y1(x) + v2(x)y2(x)
B. Examples
1. Example
d2y
dx2+ y = tan x
Answer. yc(x) = c1 sin x+ c2 cosx. yp(x) = v1(x) sin x+ v2(x) cosx.
v′1 sin x +v′2 cos x = 0
v′1 cosx −v′2 sin x =tanx
1
Thus we have v′1 = sin x, v′2 = − sin x tanx = cosx− sec x.
∫
sec x dx =
∫
sec x · sec x+ tanx
sec x+ tanxdx
=
∫
sec2 x+ sec x tanx
sec x+ tan xdx
=
∫
1
sec x+ tan xd(sec x+ tanx)
= ln |sec x+ tanx|+ C
92 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
v1(x) =∫
sin x dx v2(x) =∫
cosx− sec x dx
= − cosx+ c3 = sin x− ln |sec x+ tanx|+ c4
yp(x) = (− cos x+ c3) sin x+ (sin x− ln | sec x+ tan x|+ c4) cosx
= c3 sin x+ c4 cos x− cosx ln | sec x+ tanx|y(x) = yc(x) + yp(x)
= c1 sin x+ c2 cos x− cosx ln | sec x+ tanx|
2. Higher Order Method
Consider the general nth order linear ODE
an(x)dny
dxn+ · · ·+ a1(x)
dy
dx+ a0(x)y = F (x). (4.4.3)
The corresponding homogeneous ODE is
an(x)dny
dxn+ · · ·+ a1(x)
dy
dx+ a0(x)y = 0. (4.4.4)
The complementary function of (4.4.4) is
c1y1(x) + c2y2(x) + · · ·+ cnyn(x).
The method is to determine
yp(x) = v1(x)y1(x) + v2(x)y2(x) + · · ·+ vn(x)yn(x)
to be a particular solution of (4.4.3). The method of variation of parameters is to solve the following
system:
v′1y1(x) +v′2y2(x) + · · · +v′nyn(x) = 0
v′1y′1(x) +v′2y
′2(x) + · · · +v′ny
′n(x) = 0
...
v′1y(n−1)1 (x) +v′2y
(n−1)2 (x) + · · · +v′ny
(n−1)n (x) =
F (x)
an(x)
3. Example
d3y
dx3− 6
d2y
dx2+ 11
dy
dx− 6y = ex
4.5 The Cauchy-Euler Equation 93
Answer. yc(x) = c1ex + c2e
2x + c3e3x. yp(x) = v1(x)e
x + v2(x)e2x + v3(x)e
3x.
v′1ex +v′2e
2x +v′3e3x = 0
v′1ex +2v′2e
2x +3v′3e3x = 0
v′1ex +4v′2e
2x +9v′3e3x = ex/1
Thus v′1 = 1/2, v′2 = −ex, v′3 = e−2x/2 and we have v1 = x/2, v2 = e−x, v3 = −e−2x/4.
yp(x) =1
2xex + e−xe2x + (−1
4)e−2xe3x
=1
2xex +
3
4ex
y(x) = yc(x) + yp(x)
= c1ex + c2e
2x + c3e3x +
1
2xex
4.5 The Cauchy-Euler Equation
A. The Equation and the Method of Solution
1. The so-called Cauchy-Euler equation is the equation that is in the form
anxn d
ny
dxn+ an−1x
n−1 dn−1y
dxn−1+ · · ·+ a1x
dy
dx+ a0y = F (x) (4.5.1)
where a0, a1, . . . , an are constants and an is not identically zero .
2. Theorem 4.5.1
The transformation x = et (i.e., t = ln x) reduces (4.5.1) to a linear ODE with constant coefficients.
Example
a2x2 d
2y
dx2+a1x
dy
dx+a0y = F (x)
−→ A2d2y
dt2+A1
dy
dt+A0y = G(x)
where A2 = a2, A1 = a1 − a2, A0 = a0 and G(t) = F (et).
dy
dx=
dy
dt
dt
dx=
1
x
dy
dtd2y
dx2=
d
dx(dy
dx) =
d
dx(1
x
dy
dt)
= − 1
x2
dy
dt+
1
x
d
dx(dy
dt)
= − 1
x2
dy
dt+
1
x
d2y
dt2dt
dx= − 1
x2
dy
dt+
1
x2
d2y
dt2
94 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
B. Some Examples
1. Example
x2 d2y
dx2− 2x
dy
dx+ 2y = x3
Answer. Let x = et, t = ln x. We have A2 = 1, A1 = −2 − 1 = −3, A0 = 2 and G(t) = F (et) = e3t.
Therefore
d2y
dt2− 3
dy
dt+ 2y = e3t
Consider the characteristic equation m2 − 3m + 2 = 0. We have yc = c1et + c2e
2t. Using the UC
method we set yp = Ae3t since yc 6∈ S = e3t.
9Ae3t − 9Ae3t + 2Ae3t = e3t, A = 1/2
y(t) = c1et + c2e
2t +1
2e3t
Since et = x, we find
y(x) = c1x+ c2x2 +
1
2x3.
2. Four-Steps Procedure
Four-Steps procedure to determine the coefficients of transformation from dny/dxn.
(1) Determine r(r − 1)(r − 2) · · · [r − (n− 1)].
(2) Expand the preceding as a polynomial.
(3) Replace rk by dky/dtk, k = 1, 2, . . . , n.
(4) Equate xn dny
dxnto the result in (3).
3. Example: n=3
(1) r(r − 1)(r − 2). (2) r3 − 3r2 + 2r. (3)d3y
dt3− 3
d2y
dt2+ 2
dy
dt.
(4) x3 d3y
dx3=
d3y
dt3− 3
d2y
dt2+ 2
dy
dt.
4. Example
x3 d3y
dx3− 4x2 d
2y
dx2+ 8x
dy
dx− 8y = 4 lnx
4.6 Linear Differential Equations and Difference Equations 95
Answer. Let x = et, t = ln x. We have
xdy
dx=
dy
dt
x2 d2y
dx2=
d2y
dt2− dy
dt
x3 d3y
dx3=
d3y
dt3− 3
d2y
dt2+ 2
dy
dt
Therefore the original ODE is transferred to
d3y
dt3− 7
d2y
dt2+ 14
dy
dt− 8y = 4t
Consider the characteristic equation m3 − 7m2 + 14m − 8 = 0. We have yc = c1et + c2e
2t + c3e4t.
Using the UC method we set yp = At +B since yc 6∈ S.
14A− 8At− 8B = 4t, A = −1/2, B = −7/8
y(t) = c1et + c2e
2t + c3e4t − 1
2t− 7
8
y(x) = c1x+ c2x2 + c3x
4 − 1
2ln x− 7
8.
4.6 Linear Differential Equations and Difference Equations
A. First Order Linear Differential Equations and Difference Equations
1. Homogeneous Differential Equations
Consider
a1dy
dx+ a0y = 0 (4.6.1)
where a1 and a0 are real constants. The polynomial equation in the unknown m:
a1 ·m1 + a0 = 0 (4.6.2)
is called the characteristic equation or the auxiliary equation of the equation (4.6.1). Solve (4.6.2)
and we have m = −a0/a1. Thus the general solution of (4.6.1) is
y(x) = c · emx = c · e−a0a1
·x
where c is an arbitrary constant.
Remark
96 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
The equation (4.6.1) is separable. Thus we can solve it by the method of separation variables.
1
ydy = −a0
a1dx
ln |y| = −a0a1
x+ c1
y(x) = c · exp(−a0a1· x)
2. Nonhomogeneous Differential Equations
Consider
a1dy
dx+ a0y = F (x) (4.6.3)
where a1 and a0 are real constants.
(1) The general solution of (4.6.1) is called the complementary function of (4.6.3). We shall denote
this by yc.
(2) Any particular solution of (4.6.3) involving no arbitrary constant is called a particular integral
of (4.6.3). We shall denote this by yp.
(3) The solution yc + yp of (4.6.3) is called the general solution of (4.6.3).
3. Exampledy
dx+ y = x
Answer. (1) Complementary function yc:
Consider the char. eqn. m+ 1 = 0. Thus m = −1.
yc(x) = c · e−x.
(2) Particular integral yp (UC method):
Let yp(x) = Ax+B. Thus A+ Ax+B = x. We have A = 1 and B = −1.
yp(x) = x− 1
(3) The general solution is
y(x) = yp(x) + yc(x) = x− 1 + c · e−x
Remark: This equation is linear and we can consider the integrating factor
µ(x) = exp
(∫
1 · dx)
= ex.
Thus we can mutiply the ODE by µ(x) and have
ex · dydx
+ ex · y = ex · xd
dx(ex · y) = ex · x
ex · y(x) = xex − ex + c
y(x) = x− 1 + c · e−x
4.6 Linear Differential Equations and Difference Equations 97
4. Homogeneous Difference Equations
Consider
a1yn+1 + a0yn = 0 (4.6.4)
where a1 and a0 are real constants. The polynomial equation in the unknown m:
a1 ·m1 + a0 = 0 (4.6.5)
is called the characteristic equation or the auxiliary equation of the equation (4.6.4). Solve (4.6.5)
and we have m = −a0/a1. Thus the general solution of (4.6.4) is
yn = c ·mn = c ·(
−a0a1
)n
where c is an arbitrary constant.
Remark
a1yn+1 + a0yn = a1c
(
−a0a1
)n+1
+ a0c
(
−a0a1
)n
= c
(
−a0a1
)n(
−a1(a0a1
) + a0
)
= 0
If the initial value y0 is given then we can obtain c = y0.
We can also use the technique of recursive multiplication to solve it.
y1 =
(
−a0a1
)
y0
y2 =
(
−a0a1
)
y1
· · ·yn =
(
−a0a1
)
yn−1
Multiply these equations and we have yn =(
−a0a1
)n
· y0.5. Nonhomogeneous Difference Equations
Consider
a1yn+1 + a0yn = F (n) (4.6.6)
where a1 and a0 are real constants.
(1) The complementary solution of (4.6.6) is denoted by y(c)n .
(2) Any particular solution of (4.6.6) is denoted by y(p)n .
(3) The solution y(c)n + y
(p)n of (4.6.6) is called the general solution of (4.6.6).
6. Example
yn+1 − 2yn = 3n− 5
98 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
Answer. (1) Complementary solution y(c)n :
Consider the char. eqn. m− 2 = 0. Thus m = 2.
y(c)n = c · 2n.
(2) Particular solution y(p)n (UC method):
Let y(p)n = An +B. Thus A(n+ 1) +B − 2(An+B) = 3n− 5. We have A = −3 and B = 2.
y(p)n = −3n+ 2
(3) The general solution is
yn = y(c)n + y(p)n = −3n + 2 + c · 2n
Remark: if we have the initial value y0 then c = y0 − 2.
B. High-Order Differential and Difference Equations
1. Homogeneous and Nonhomogeneous Differential Equations
Consider the nth-order nonhomogeneous linear ODE (4.1.1) and the corresponding homogeneous
equation (4.1.2).
(1) The general solution of (4.1.2) is called the complementary function of (4.1.2). We shall denote
this by yc.
(2) Any particular solution of (4.1.1) involving no arbitrary constant is called a particular integral
of (4.1.1). We shall denote this by yp.
(3) The solution yc + yp of (4.1.1) is called the general solution of (4.1.1).
2. Homogeneous Difference Equations
Consider the nth-order homogeneous linear difference equation
anyn + an−1yn−1 + · · ·+ an−ryn−r = 0 (4.6.7)
where an, an−1, . . . , an−r are real constants. The polynomial equation in the unknown m:
anmr + an−1m
r−1 + · · ·+ an−r+1m+ an−r = 0 (4.6.8)
is called the characteristic equation or the auxiliary equation of the equation (4.6.7).
Case 1. Distinct Real Roots
Theorem 4.6.1
Consider the equation (4.6.7) with constant coefficients. If the characteristic equation (4.6.8) has the
r distinct real roots m1, m2, . . . , mr, then the general solution of (4.6.7) is
yn = c1mn1 + c2m
n2 + · · ·+ crm
nr
4.6 Linear Differential Equations and Difference Equations 99
where c1, c2, . . . , cr are arbitrary constants.
Example
yn − 6yn−1 + 8yn−2 = 0, n ≥ 2
Answer. The char. eqn. is m2 − 6m+ 8 = 0. Thus m = 2 and 4.
yn = c14n + c22
n.
If the initial values are y0 = 0 and y1 = 2. Then we have c1 = 1 and c2 = −1. So yn = 4n − 2n.
Case 2. Repeated Real Roots
Theorem 4.6.2
Consider the equation (4.6.7) with constant coefficients. Assume that the characteristic equation
(4.6.8) has the s distinct real roots m1, m2, . . . , ms and s ≤ r. If the multiplicity of mi is ω(i),
i = 1, · · · , s then
for the root mi we have the solution
µi(n) =(
c(i)0 + c
(i)1 n+ c
(i)2 n2 + · · ·+ c
(i)
ω(i)−2nω(i)−2 + c
(i)
ω(i)−1nω(i)−1
)
mni
and the general solution of (4.6.7) is
yn = µ1(n) + µ2(n) + · · ·+ µs(n)
where c(i)0 , c
(i)1 , . . . , c
(i)
ω(i)−1are arbitrary constants.
Example
yn − 7yn−1 + 16yn−2 − 12yn−3 = 0, n ≥ 3
Answer. The char. eqn. is m3 − 7m2 + 16m− 12 = 0. Thus m = 2, 2 and 3.
yn = (c(1)0 + c
(1)1 n)2n + c
(2)0 3n.
Simply we may write yn = (c0 + c1n)2n + c23
n.
Case 3. Conjugate Complex Roots
Theorem 4.6.3
Consider the equation (4.6.7) with constant coefficients. Assume that the characteristic equation
(4.6.8) has the two conjugate complex roots m1 = reiθ and m2 = re−iθ (in polar form) If the
multiplicity of mi is 1 then
for the root m1 and m2 we have the solution
yn = C1rn cos(nθ) + C2r
n sin(nθ).
100 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
c1mn1 + c2m
n2 = c1r
neinθ + c2rne−inθ
= rn[c1(cosnθ + i sin nθ) + c2(cosnθ − i sinnθ)
= C1rn cos(nθ) + C2r
n sin(nθ),
where C1 = c1 + c2 and C2 = i(c1 − c2).
Example
yn+2 − 2yn+1 + 2yn = 0
Answer. The char. eqn. is m2− 2m+2 = 0. Thus m1,2 = 1± i. Hence r =√2 and θ = π/4, so that
the general solution is
yn = (√2)n(
C1 cosnπ
4+ C2 sin
nπ
4
)
.
3. Nonhomogeneous Difference Equations
Consider the nth-order homogeneous linear difference equation
anyn + an−1yn−1 + · · ·+ an−ryn−r = F (n) (4.6.9)
where an, an−1, . . . , an−r are real constants.
(1) The complementary solution of (4.6.9) is denoted by y(c)n .
(2) Any particular solution of (4.6.9) is denoted by y(p)n .
(3) The solution y(c)n + y
(p)n of (4.6.9) is called the general solution of (4.6.9).
Case 1. F (n) is a polynomial of n
Let F (n) = b0 + b1n+ · · ·+ bknk. We can use UC method. Let
y(p)n = A0 + A1n+ · · ·+ Ak+pnk+p
where
p = 0 if 1 is not a root of the char. eqn.
p = ω if 1 is a root of the char. eqn. and ω is the multiplicity of 1
Actually, we can use
y(p)n = Aωnω + Aω+1n
ω+1 + · · ·+ Ak+ωnk+ω.
Remark (Special Case, F (n) = b0)
In the special case when F (n) = b0 is a constant, a particular solution can easily be obtained by
setting y(p)n = A. Substitution of yn = A in the original difference equation leads to the determination
A =b0
an + an−1 + · · ·+ an−r
4.6 Linear Differential Equations and Difference Equations 101
provided that the sum of the coefficients does not vanish.
Example
yn+2 − 2yn+1 + 2yn = 1
Answer. A particular solution can be found by y(p)n = 1/(1 − 2 + 2) = 1. The char. eqn. is
m2 − 2m+ 2 = 0. Thus m1,2 = 1± i. Hence r =√2 and θ = π/4, so that the general solution is
yn = y(p)n + y(c)n = 1 + (√2)n(
C1 cosnπ
4+ C2 sin
nπ
4
)
.
Example
yn − 7yn−1 + 10yn−2 = 4n− 5, n ≥ 2
Answer. (1) Complementary solution: the char. eqn. is m2 − 7m+ 10 = 0. Thus m = 2 and 5.
y(c)n = c12n + c25
n
(2) Particular solution: we can consider
y(p)n = A+Bn
because 1 is not a root of the char. eqn. So we have
(A+Bn)− 7(A+Bn− B) + 10(A+Bn− 2B) = 4n− 5.
Thus, A = 2 and B = 1 and y(p)n = 2 + n. The general solution is
yn = y(c)n + y(p)n = c12n + c25
n + n+ 2.
If the initial values are y0 = 2 and y1 = 4 then we have the solution yn = 5 · 2n − 5n + n− 2.
Example
yn − 2yn−1 + yn−2 = 2, n ≥ 2, y0 = 1, y1 = 1
Answer. (1) Complementary solution: the char. eqn. is m2 − 2m+ 1 = 0. Thus m = 1 and 1.
y(c)n = c1 + c2n
(2) Particular solution: Since the multiplicity of root 1 is that ω = 2, we can consider
y(p)n = An2
102 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
So we have
0− 2A+ 4A = 2.
Thus, A = 1 and y(p)n = n2. The general solution is
yn = y(c)n + y(p)n = c1 + c2n + n2.
Since the initial values are y0 = 1 and y1 = 1, we have c1 = 1 and c2 = −1. The solution is
yn = 1− n + n2.
Case 2. F (n) is an exponential function of n
Let F (n) = b · dn where b and d are constants. We can use UC method. Let
y(p)n = (A0 + A1n+ · · ·+ Apnp) dn
where
p = 0 if d is not a root of the char. eqn.
p = ω if d is a root of the char. eqn. and ω is the multiplicity of d
Example
yn + yn−1 − 2yn−2 = 3 · 2n, n ≥ 2
Answer. (1) Complementary solution: the char. eqn. is m2 +m− 2 = 0. Thus m = −2 and 1.
y(c)n = c1(−2)n + c2
(2) Particular solution: we can consider
y(p)n = A · 2n
because 2 is not a root of the char. eqn. So we have
A · 2n + A · 2n−1 − 2A · 2n−2 = 3 · 2n.
Thus, A = 3 and y(p)n = 3 · 2n. The general solution is
yn = y(c)n + y(p)n = 3 · 2n + c1(−2)n + c2.
Exercise
yn − 4yn−1 + 3yn−2 = 2 · 3n, n ≥ 2
Remark: Nonphysical Oscillations
Numerical difficulties with boundary layer problems.
4.6 Linear Differential Equations and Difference Equations 103
• It may fail to convergence. (Typical iteration schemes)
• It may exhibit oscillations which are physical unrealistic.
• If an explicit time-stepping procedure is used, it may become unstable for unexpectedly small
time steps.
A simple example
−ǫ · u′′ + u′ = 0, on (0, 1), ǫ > 0,
u(0) = 0, u(1) = 1.
The characteristic equation is
−ǫλ2 + λ = 0
We have u(x) = c1e0 + c2e
x/ǫ. Initial conditions imply that c1 = −1/(e1/ǫ − 1) and c2 = 1/(e1/ǫ − 1).
u(x) =ex/ǫ − 1
e1/ǫ − 1
The standard finite element method (FEM) or central difference scheme is
−ǫ · uj+1 − 2uj + uj−1
h2+
uj+1 − uj−1
2h= 0, j = 1, 2, · · · , N − 1
u0 = 0, uN = 1.
Define the number, mesh Peclet number,
βdef=
h
ǫ> 0.
Rewrite the equation and we have
(1
2β − 1)uj+1 + 2uj + (−1
2β − 1)uj−1 = 0
The characteristic equation is
(1
2β − 1)λ2 + 2λ+ (−1
2β − 1) = 0
We have uh(xj) = c1λj1 + c2λ
j2 where λ1 = 1 and the important char. root λ2
λ2 =1 + 0.5β
1− 0.5β.
104 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE
0 0.2 0.4 0.6 0.8 1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
X−Axis
Y−
Axi
s
9 mesh points17 mesh points65 mesh pointsExact solution
ε=0.01
Initial conditions imply that c1 = −1/(λN2 − 1) and c2 = 1/(λN
2 − 1) and we have
uh(xj) =λj2 − 1
λN2 − 1
.
It is clear that the approximate solution is oscillatory when β > 2, i.e., λ2 < 0.
References:
1. Elementary Numerical Analysis, Samuel D. Conte and Carl de Boor, 1987
2. Lecture Note, Prof. Ming-Luen Wu
3. Numerical Solution of Convection-Diffusion Problems, K. W. Morton, Publisher: Chapman &
Hall/CRC; 1 edition (May 15, 1996)
4. Numerical Solution of Partial Differential Equations by the Finite Element Method, C. Johnson,
1987
Chapter 5
Applications of Second-Order Linear
ODE
5.1 The Differential Equation of Vibrations of a Mass on a
Spring
A. Mechanical System – Spring Vibration
1. Hooke’s law
The magnitude of the force needed to produce a certain elongation, provided this elongation is not
too great. In mathematical form,
|F | = ks,
where F is the magnitude of the force, s is the amount of elongation, and k is a constant of propor-
tionality which we shall call the spring constant.
When a mass is hung upon a spring of spring constant k and thus produces an elongation of amount
s, the force F of the mass upon the spring therefore has magnitude ks. The spring at the same time
exerts a force upon the mass called the restoring force of the spring. This force is equal in magnitude
but opposite in sign to F and hence has magnitude −ks.
106 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
L
Fig. (a)
L
l
m
Fig. (b)
L
l
O
x
m
Fig. (c)
(a) natural length L
(b) mass in equilibrium position; spring has stretched length L+ l.
(c) mass distance x below equilibrium position; spring has stretched length L+ l + x.
2. Forces Acting Upon the Mass
(a) F1, the force of gravity of magnitude mg.
F1 = mg ↓ positive (5.1.1)
(b) F2, the restoring force of the spring.
F2 = −k(x+ l) ↑ negative (5.1.2)
When x = 0, F1 + F2 = 0, thus we have mg = kl.
Replacing kl by mg in (5.1.2) we see that
F2 = −kx−mg. (5.1.3)
(c) F3, the resisting force of the medium, called the damping force. For small velocities it is ap-
proximately proportional to the magnitude of the velocity:
|F3| = a
∣
∣
∣
∣
dx
dt
∣
∣
∣
∣
(5.1.4)
where a > 0 is called the damping constant. When the mass is moving downward, F3 acts in
the upward direction and so F3 < 0.
F3 = −adx
dt, (a > 0) ↑ negative (5.1.5)
5.2 Free, Undamped Motion 107
(d) F4, any external impressed forces that act upon the mass.
F4 = F (t) (5.1.6)
We now apply Newton’s second law, F = ma, where F = F1 + F2 + F3 + F4. We find
md2x
dt2= mg − kx−mg − a
dx
dt+ F (t) (5.1.7)
or
md2x
dt2+ a
dx
dt+ kx = F (t) (5.1.8)
(e) It is a nonhomogeneous second-order linear differential equation with constant coefficients.
(f) a = 0: If a = 0, the motion is called undamped; otherwise (a 6= 0)it is called damped.
(g) F (t) = 0: If F (t) = 0, the motion is called free; otherwise it is called forced.
5.2 Free, Undamped Motion
1. Free, Undamped Motion
In this section we consider the free, undamped motion. a = 0 and F (x) = 0. The differential
equation (5.1.8) then reduces to
md2x
dt2+ kx = 0 (5.2.1)
where m(> 0) is the mass and k(> 0) is the spring constant. Let λ2 =k
m, we rewrite (5.2.1) in the
formd2x
dt2+ λ2x = 0 (5.2.2)
Hence the general solution of (5.2.2) can be written
x = c1 sinλt + c2 cosλt (5.2.3)
where c1 and c2 are arbitrary constants.
Initial conditions:
x(0) = x0, (initial displacement) (5.2.4)
x′(0) = v0, (initial velocity) (5.2.5)
Therefore
dx/dt|t=0 = x′(0) = v0 =⇒ [c1λ cosλt− c2λ sinλt]|t=0 = v0 (5.2.6)
108 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
We have
c2 = x0 and c1 =v0λ
The solution is
x(t) =v0λsin λt+ x0 cosλt (5.2.7)
An alternative form
x(t) = c[v0/λ
csin λt+
x0
ccos λt]
where c =√
(v0/λ)2 + x02. Let φ be defined by satisfying
v0/λ
c= − sin φ and
x0
c= cosφ. So
x(t) = c cos(λt+ φ) (5.2.8)
=√
(v0/λ)2 + x02 cos
(√
k
mt + φ
)
(5.2.9)
2. Simple Harmonic Motion
(1) The free, undamped motion of the mass is a simple harmonic motion.
(2) The constant c is called the amplitude of the motion and gives the maximum (positive) displace-
ment of the mass from its equilibrium position.
(3) The motion is a periodic motion.
x = c⇔√
k
mt+ φ = ±2nπ, n ∈ N ∪ 0
Therefore
t =
√
m
k(±2nπ − φ) > 0 (5.2.10)
The time interval between two successive maxima is called the period of the motion and it is given
by2π
λ=
2π√
k/m= 2π ·
√
m
k. (5.2.11)
(4) The reciprocal of the period, which gives the number of oscillations per second, is called frequency
( or natural frequency ) of the motion and it is given by
λ
2π=
√
k/m
2π=
1
2π·√
k
m. (5.2.12)
(5) The number of φ is called the phase constant (or phase angle). See Fig. 5.1.
3. Example
An 8-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight
comes to rest in its equilibrium position, thereby stretching the spring 6 in. The weight is then pulled
5.2 Free, Undamped Motion 109
0 0.5 1 1.5 2
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
π /8 2π/8 3π/8 4π/8
Period π /4
Amplitude 0.28
t = −φ/8
X
t
Figure 5.1: x(t) = 0.28 cos(8t + φ) where φ=-0.46
down 3 in. below its equilibrium position and released at t = 0 with an initial velocity of 1 ft/sec,
directed downward. Neglecting the resistance of the medium and assuming that no external forces
are present, determine the amplitude, period and frequency of the resulting motion.
(1) Formulation.
m = w/g = 8/32 (slugs) 6 in. = 1/2 ft.
F = ks ⇒ 8 = k/2 k = 16 (lb/ft)
8/32d2x/dt2 + 16x = 0 d2x/dt2 + 64x = 0
x(0) = x0 = 3/12 = 1/4 (ft) x′(0) = v0 = 1 (ft/sec)
The differential equation is
d2x
dt2+ 64x = 0
x(0) =1
4, x′(0) = 1
110 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
(2) Solution.
m2 + 64 = 0 m = 0± 8i
x(t) = c1 sin 8t+ c2 cos 8t x(0) = 1/4, x′(0) = 1⇒ c1 = 1/8, c2 = 1/4
x(t) =1
8sin 8t+
1
4cos 8t c =
√
(
1
8
)2
+
(
1
4
)2
=
√5
8
=
√5
8
(√5
5sin 8t+
2√5
5cos 8t
)
=
√5
8cos(8t+ φ) φ
.= −0.46 (radians)
The solution is
x(t).= 0.280 cos(8t− 0.46)
The amplitude of the motion is c =√58
.= 0.280 (ft). The period is 2π/8 = π/4 (sec). The frequency
is 4/π oscillations/sec. The graph is shown in Fig. 5.1.
5.3 Free, Damped Motion
We now consider the effect of the resistance of the medium upon the mass on the spring. Still
assuming that no external forces are present.
md2x
dt2+ a
dx
dt+ kx = 0 (5.3.1)
x(0) = x0, x′0(0) = v0 (5.3.2)
Let λ2 = k/m and 2b = a/m (for convenience). We have
d2x
dt2+ 2b
dx
dt+ λ2x = 0 (5.3.3)
The characteristic equation is
m2 + 2bm+ λ2 = 0.
Thus m = −b ±√
b2 − λ2.
b2 − λ2 =a2
4m2− k
m=
a2 − 4mk
4m2
Case 1. Underdamping case, b2 − λ2 < 0 (i.e. a2 − 4mk < 0): Complex conjugate roots. Damped,
oscillatory motion.
Case 2. Critical damping case, b2 − λ2 = 0 (i.e. a2 − 4mk = 0): A real double root.
Case 3. Overdamping case, b2−λ2 > 0 (i.e. a2−4mk > 0): Distinct real roots. Overcritical damping
5.3 Free, Damped Motion 111
motion.
Let us discuss these three cases separately.
Case 1. Underdamping Motion
m = −b±√
b2 − λ2 = −b±√
λ2 − b2 i
x(t) = e−bt(c1 sin√
λ2 − b2 t + c2 cos√
λ2 − b2 t)
Let c =√
c21 + c22 > 0 and φ is determined by
c1√
c21 + c22= − sinφ
c2√
c21 + c22= +cosφ.
We have
x(t) = ce−bt cos(√
λ2 − b2 t + φ). (5.3.4)
Notice that
(1) The factor ce−bt is called the damping factor, or time-varying amplitude. c > 0 and b > 0 ⇒ce−bt → 0 as t→∞.
(2) cos(√
λ2 − b2 t+ φ) is of a period, oscillatory character.
(3) The oscillations are said to be ”damped out” because ce−bt → 0 as t → ∞. The motion is
described as damped, oscillatory motion.
(4) The time interval between two successive (positive) maximum displacement is still referred to
as the period.
(5) The period is2π
√
λ2 − b2
because of the oscillatory term, cos(√
λ2 − b2 t+ φ).
(6) The ratio of the amplitude at t = T and t = T −(
2π√
λ2 − b2
)
, one period before T , is the
constant
exp
(
− 2πb√
λ2 − b2
)
.
(7) The quantity
(
2πb√
λ2 − b2
)
is the decrease in the logarithm of the amplitude ce−bt over a time
interval of one period. It is called the logarithmic decrement.
112 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
0 1 2 3 4 5 6
−0.6
−0.4
−0.2
0
0.2
0.4
0.6 Period
+ce−bt
−ce−bt
1.0 2.0 3.0 4.0 5.0
Figure 5.2: Damped, oscillatory motion.
Now we consider the initial conditions (5.3.2).
x(t) = e−bt(c1 sin√
λ2 − b2 t+ c2 cos√
λ2 − b2 t) = ce−bt cos(√
λ2 − b2 t+ φ)
x(0) = x0 = c1 · 0 + c2 · 1 ⇒ c2 = x0
x′(0) = v0 = −b · c2 + c1√
λ2 − b2 ⇒ c1 =v0 + bx0√
λ2 − b2
Let c =√
c21 + c22 and we have
x(t) = ce−(a/2m)t cos
(√
k
m− a2
4m2t + φ
)
(5.3.5)
Notice that
(1) If damping were not present, a would be equal to 0.
x(t) = c cos
(√
k
m− 0 t + φ
)
where c1 = v0/λ =√mv0/
√k, c2 = x0 and c =
√
mv20/k + x20. Moreover,
cosφ =x0
cand sinφ = −
√
mv20/k
c.
The solution (5.3.5) of damped, oscillatory motion will reduce to the solution (5.2.9) of un-
damped, free motion.
5.3 Free, Damped Motion 113
(2) b2−λ2 < 0 ⇒ b < λ ⇒ a/2m <√
k/m ⇒ a <√2km. We can say that damped, oscillatory
motion occurs when
a <√2km. (5.3.6)
(3) The frequency of the oscillations cos
(√
k
m− a2
4m2t+ φ
)
is
1
2π
√
k
m− a2
4m2. (5.3.7)
(4) The frequency (5.3.7) of the oscillations is less than the natural frequency (5.2.12),1
2π
√
k
m, of
the corresponding undamped system.
Case 2. Critical Damping Motion
x(t) = (c1 + c2t)e−bt.
Notice that
(1) The motion is no longer oscillatory.
(2) Any slight decrease in the amount of damping will change the situation back to that of Case 1.
The motion in Case 2 is said to be critically damped.
(3) x(t)→ 0 as t→∞.
Case 3. Overdamping Motion
The characteristic equation is m2 + 2bm + λ2 = 0. Therefore m1 = −b +√
b2 − λ2 and m2 =
−b−√
b2 − λ2. So
x(t) = c1em1t + c2e
m2t.
Notice that
(1) No oscillations can occur.
(2) We can no longer say that every decrease in the amount of damping will result in oscillations,
as we could in Case 2. The motion in Case 3 is said to be overcritically damped ( or simply
overdamped).
(3) x(t)→ 0 as t→∞.
114 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
Example
An 32-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight
comes to rest in its equilibrium position, thereby stretching the spring 2 ft. The weight is then pulled
down 6 in. below its equilibrium position and released at t = 0 without any initial velocity. Assuming
that no external forces are present, determine the amplitude, period and frequency of the resulting
motion with the following resistance of the medium where dx/dt is the instantaneous velocity in feet
per second and a is the damping constant.
(a) 4 (dx/dt) (a= 4) (b) 8 (dx/dt) (a= 8) (c) 10 (dx/dt) (a= 10).
Formulation.
m = w/g = 32/32 = 1 (slug). 32 = k 2 and so k = 16 lb/ft.
Equations (a), (b) and (c) are
(a)d2x
dt2+ 4
dx
dt+ 16x = 0
(b)d2x
dt2+ 8
dx
dt+ 16x = 0
(c)d2x
dt2+ 10
dx
dt+ 16x = 0.
The initial conditions are
x(0) = 1/2, x′(0) = 0.
Solution.
(a) The auxiliary equation is m2+4m+16 = 0. Its roots are conjugate complex numbers −2±2√3i.
The general solution is
x(t) = e−2t(c1 sin 2√3t+ c2 cos 2
√3t).
ICs imply c1 =√3/6 and c2 = 1/2. The solution is
x(t) = e−2t
(√3
6sin 2√3t +
1
2cos 2√3t
)
=
√3
3e−2t cos
(
2√3t− π
6
)
Interpretation.
This is a damping oscillatory motion.
damping factor
√3
3e−2t
period 2π/2√3 =√3π/3
logarithmic decrement 2√3π/3
5.4 Forced Motion 115
(b) The auxiliary equation is m2 +8m+16 = 0. Its equal roots are −4, −4. The general solution is
x(t) = (c1 + c2t)e−4t.
ICs imply c1 = 1/2 and c2 = 2. The solution is
x(t) =
(
1
2+ 2t
)
e−4t
Interpretation.
This is a critically damped motion.
(c) The auxiliary equation is m2 + 10m+ 16 = 0. Its roots are −2, −8. The general solution is
x(t) = c1e−2t + c2e
−8t.
ICs imply c1 = 2/3 and c2 = −1/6. The solution is
x(t) =2
3e−2t − 1
6e−8t
Interpretation.
This is a overdamped motion.
5.4 Forced Motion
In this section we consider the model
md2x
dt2+ a
dx
dt+ kx = F (t) (5.4.1)
where F (t) is called the external impressed force, input force or driving force. A corresponding
solution is called an output or a response of the system to the driving force.
Of particular interest are periodic inputs. We shall consider a sinusoidal input, say,
F (t) = F1 cosωt
where F1 > 0 and ω are constants. Then (5.4.1) becomes
md2x
dt2+ a
dx
dt+ kx = F1 cosωt (5.4.2)
Dividing through by m and letting
a
m= 2b,
k
m= λ2,
F1
m= E1,
116 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
this takes the more convenient form
d2x
dt2+ 2b
dx
dt+ λ2x = E1 cosωt (5.4.3)
Of particular interest is that the damping is less than critical. We shall assume that
b < λ (i.e. a < 2√km).
Hence by (5.2.4) the complementary function of (5.4.3) can be written
xc(t) = ce−bt cos(√
λ2 − b2 t+ φ). (5.4.4)
We shall now find a particular integral by the method of undetermined coefficients. Let
xp(t) = A cosωt+B sinωt.
Then we have
dxp
dt= −ωA sinωt+ ωB cosωt,
d2xp
dt2= −ω2A cosωt− ω2B sinωt
−2bωA+ (λ2 − ω2)B = 0, (λ2 − ω2)A+ 2bωB = E1
A =E1(λ
2 − ω2)
(λ2 − ω2)2 + 4b2ω2, B =
2bωE1
(λ2 − ω2)2 + 4b2ω2.
The particular integral is
xp(t) =E1
(λ2 − ω2)2 + 4b2ω2
[
(λ2 − ω2) cosωt+ 2bω sinωt]
(5.4.5)
We now put this in the alternative ”phase angle” form. Let
cos θ =λ2 − ω2
√
(λ2 − ω2)2 + 4b2ω2
sin θ =2bω
√
(λ2 − ω2)2 + 4b2ω2
,(5.4.6)
and we have
xp(t) =E1
√
(λ2 − ω2)2 + 4b2ω2
cos(ωt− θ). (5.4.7)
The general solution of (5.4.2) is
x(t) = xc(t) + xp(t) = ce−bt cos(√
λ2 − b2 t+ φ) (5.4.8)
+E1
√
(λ2 − ω2)2 + 4b2ω2
cos(ωt− θ).
Observe that
5.5 Resonance Phenomena 117
(1) The first term, ce−bt cos(√
λ2 − b2 t + φ), represents the damped oscillation. If F (t) = 0, then
(5.4.8) reduce to (5.3.4).
(2) t→∞ ⇒ the first term → 0. The first term is thus called the transient term.
(3) t→∞ ⇒ x(t)→ the second term . The second term is thus called the steady-state term.
5.5 Resonance Phenomena
Define the function f of ω by
f(ω) =E1
√
(λ2 − ω2)2 + 4b2ω2
(5.5.1)
We know that
ω = 0 ⇒ f(0) =E1
λ2 and F (t) is the constant F1.
ω →∞ ⇒ f(ω)→ 0.
Let us consider 0 < ω <∞ and find the maximum of f(ω).
f ′(ω) = 0 ⇒ 4ω[2b2 − (λ2 − ω2)] = 0 ⇒ ω = 0 or ω =√
λ2 − 2b2.
(1) λ2 < 2b2:√
λ2 − 2b2 is a complex number. f ↓ 0. f has no extremum.
(2) λ2 = 2b2: f(ω) =E1
√
λ4 + ω4. f ↓ 0. f has no extremum.
(3) λ2 > 2b2: (i.e. a <√2km ) f(ω) has a relative maximum at ω1 =
√
λ2 − 2b2.
f(ω1) =E1
2b√
λ2 − b2.
Return to the original notation. We have
f(ω) =F1/m
√
(
k
m− ω2
)2
+( a
m
)2
ω2
(5.5.2)
f(ω) has a relative maximum at ω1 =√
λ2 − 2b2 =
√
k
m− a2
2m2.
f(ω1) =F1/m
a
m
√
k
m−( a
2m
)2.
118 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
If ω = ω1 then the forcing function is said to be in resonance with the system. The value ω1/2π is
called the resonance frequency of the system. For the special case m = k = E1 = 1,
f(ω) =1
√
ω4 + (−2 + a2)ω2 + 1.
See Fig. 5.3 for the relation between f(ω) and ω when a is approximated to 0. The graph of f(ω)
0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
2
2.5
3
3.5
4
4.5
a=1.5
a=1.25
a=1.0
a=0.5
a=0.25
ω
f(ω
)
Figure 5.3: The relation between f(ω) and ω when a is approximated to 0.
is called the resonance curve of the system. For a given system with fixed m, k, and F1, there is a
resonance curve corresponding to each value of the damping coefficient a ≥ 0.
Notice that
(a) Resonance ←→ Max. amplitude.
(b) Resonance frequency
ω1
2π=
1
2π
√
k
m− a2
2m2. (5.5.3)
5.5 Resonance Phenomena 119
Frequency Formula and notation
Natural frequencyω0
2πdef=
1
2π
√
k
m
ω0 = λ =
√
k
m
Resonance frequencyω1
2πdef=
1
2π
√
k
m− a2
2m2
ω1 =√
λ2 − 2b2 =
√
k
m− a2
2m2
The corresponding free, damped frequencyωfd
2πdef=
1
2π
√
k
m− a2
4m2
ωfd =√
λ2 − b2 =
√
k
m− a2
4m2
The special case a = 0 will imply ω0 = ω1 = ωfd.
(c) Resonance can occur only if λ2 > 2b2. The underdamping case can occur only if λ2 > b2. Since
2b2 < λ2 ⇒ b2 < λ2, the damping must be less than critical in such a case.
(d) The resonance frequency is less than of the corresponding free, damped oscillation.
ω1
2π=
1
2π
√
λ2 − 2b2 <ωfd
2π=
1
2π
√
λ2 − b2
ω1
2π=
1
2π
√
k
m− a2
2m2<
ωfd
2π=
1
2π
√
k
m− a2
4m2
Now we will discuss forced motion with two case, undamped and damped cases.
Case 1. Undamped Forced Oscillation (a = 0)
md2x
dt2+ kx = F1 cosωt
120 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
(1) If ω 6= ω0, i.e. ω 6=√
k
m, then the UC set is cosωt, sinωt. From (5.4.5) we have
xp(t) =E1
(λ2 − ω2)2 + 0
[
(λ2 − ω2) cosωt+ 0]
(i.e. b = 0)
=E1
λ2 − ω2cosωt.
f(ω) =E1
λ2 − ω2=
F1/mkm− ω2
(amplitude of xp).
ω0
2π=
ω1
2π=
1
2π
√
k
m− 0 =
1
2π
√
k
mcycles/sec
= the ”natural frequency” of the undamped, free system.
xp(t) =F1/m
k/m− ω2cosωt
=F1
k· 1
1− (ω/ω0)2cosωt
ρ(ω) ≡ 1
1− (ω/ω0)2, ω → ω0 ⇒ ρ→∞
This phenomenon of excitation of large oscillations by matching input (ω) and natural frequency
(ω0) is known as resonance (ω = ω0).
Notice that
(a) The quantity ρ is called the resonance factor. See Fig. 5.4 for the special case ω0 = ω1 = 1.
(b)
ρ
k=
f(ω)
F1=
the amplitude of output (xp)
the amplitude of input
The ratio is the so-called amplification.
(2) If ω = ω0, i.e. ω =
√
k
m, then the UC set is t · cosωt, t · sinωt.
md2x
dt2+ kx = F1 cos
√
k
mt
d2x
dt2+
k
mx = E1 cos
√
k
mt
xc(t) = c1 sin
√
k
mt+ c2 cos
√
k
mt
xp(t) = At · sin√
k
mt +Bt · cos
√
k
mt
5.5 Resonance Phenomena 121
0 0.5 1 1.5 2 2.5 3−10
−8
−6
−4
−2
0
2
4
6
8
10
ω1
ω
ρ(ω
)
Figure 5.4: The relation between ρ and ω.
By the UC method we have A =E1
2
√
m
kand B = 0. The solution x(t) is that
x(t) = c1 sin
√
k
mt + c2 cos
√
k
mt +
E1
2
√
m
kt · sin
√
k
mt
= c · cos(√
k
mt+ φ
)
+E1
2
√
m
kt · sin
√
k
mt
Notice that
(a) The graph of the functionE1
2
√
m
kt · sin
√
k
mt is shown in Fig. 5.5 for the special case
m = k = E1 = 1.
(b) However, common sense intervenes and convinces us that before this exciting phenomenon
can occur the system will break down and the ODE will no longer apply!
(3) Another interesting and highly important type of oscillation is obtained when ω is closed to ω0.
ω.= ω0 but ω 6= ω0.
122 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
0 2 4 6 8 10 12 14
−6
−4
−2
0
2
4
6
t
1/2
⋅ t s
in(t
)
Figure 5.5: The particular integral xp(t) =12t sin(t).
x(t) = xc(t) + xp(t)
= c1 sinω0t + c2 cosω0t+F1/m
k
m− ω2
cosωt
= c1 sinω0t + c2 cosω0t+F1
m(ω20 − ω2)
cosωt
Initial conditions x(0) = 0 and x′(0) = 0 imply c1 = 0 and c2 = −F1
m(ω20 − ω2)
. Then we have
x(t) =F1
m(ω20 − ω2)
(cosωt− cosω0t)
=2F1
m(ω20 − ω2)
sinω0 + ω
2t · sin ω0 − ω
2t
Since ω is close to ω0, the difference ω0 − ω is small, so that the period of the last sine function is
large, and we obtain an oscillation of the type shown in Fig. 5.6 which is a solution of the following
example. It is so-called a beat.
Example (Undamped forced motion)
5.5 Resonance Phenomena 123
0 5 10 15 20 25 30 35 40
−1.5
−1
−0.5
0
0.5
1
1.5
← K sin((3−ω)t/2)
← −K sin((3−ω)t/2)
Figure 5.6: Forced undamped oscillation when the difference of the input and natural frequencies is
small (ω = 2.56).
2d2x
dt2+ 18x = 3cos(ωt), t > 0,
x(0) = 0, x′(0) = 0.
We consider two cases: (a) ω 6= 3 (b) ω = 3
(a) ω 6= 3
x(t) = xc(t) + xp(t) where
xp(t) =3
18− 2ω2cos(ωt)
xc(t) = c1 sin(3t) + c2 cos(3t)
c1 = 0, c2 =−3
18− 2ω2
x(t) = xp(t) + xc(t)
=−3
18− 2ω2cos(3t) +
3
18− 2ω2cos(ωt)
=6
18− 2ω2sin(
3 + ω
2t) sin(
3− ω
2t)
See Fig. 5.6 for the case ω = 2.65 (beat), Fig. 5.7 for ω = 5 and Fig. 5.8 for ω = 0.5.
124 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
0 5 10 15 20 25 30 35 40
−1
−0.5
0
0.5
1
ω = 5
Figure 5.7: Forced undamped oscillation, ω = 5.
0 5 10 15 20 25 30 35 40
−1
−0.5
0
0.5
1
ω = 0.5
Figure 5.8: Forced undamped oscillation, ω = 0.5.
(b) ω = 3
2d2x
dt2+ 18x = 3cos(ωt), t > 0,
x(0) = 0, x′(0) = 0.
ω = 3
We know that x(t) = xc(t) + xp(t) where
xp(t) =1
4t sin(3t)
xc(t) = c1 sin(3t) + c2 cos(3t)
c1 = 0, c2 = 0
Thus x(t) = 14t sin(3t). See Fig. 5.9 for the solution.
Case 2. Damped Forced Oscillation (a 6= 0)
md2x
dt2+ a
dx
dt+ kx = F1 cosωt
From (5.4.8) we have
x(t) = xc(t) + xp(t) = ce−bt cos(√
λ2 − b2 t+ φ) (5.5.4)
+E1
√
(λ2 − ω2)2 + 4b2ω2
cos(ωt− θ).
Notice that
(1) This is what happens in practice, because no physical system is completely undamped.
(2) The amplitude will always be finite.
(3) The amplitude f(ω) has a maximum at ω1 =
√
k
m− a2
2m2depending on a. This may be called
5.5 Resonance Phenomena 125
0 5 10 15 20 25 30 35 40−10
−8
−6
−4
−2
0
2
4
6
8
10
x(t)= t/4
x(t)= − t/4
Figure 5.9: ω = ω1
practical resonance.
(4) It is of great importance because it shows that some input may excite oscillations with such a
large amplitude that the system can be destroyed.
(5) Machines, cars, airplanes and bridges are vibrating mechanical systems, and it is sometimes
rather difficult to find constructions which are completely free of undesired resonance effects.
(6) Another expression of the amplitude of xp(t) (amplification)
From (5.5.1) and (5.5.2) we have
f(ω) =E1
√
(λ2 − ω2)2 + 4b2ω2
(5.5.5)
w20 = λ2 = k/m, ω1 =
√
k
m− a2
2m2, a2 = 2m2(ω2
0 − ω21) and
f(ω1) =F1/m
√
(ω20 − ω2
1)2 +
( a
m
)2
ω21
=F1/m
√
(
k
m− ω2
1
)2
+( a
m
)2
ω21
=F1/m
√
(ω20 − ω2
1)2 + 2(ω2
0 − ω21)ω
21
=F1/m
√
a4
4m4+
a2
m2ω21
=2mF1
a√
a2 + 4m2ω21
=2mF1
a√
2m2ω20 + 2m2ω2
1
126 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
Fig. 5.2 shows the amplification f(ω)/F1 as a function of ω for m = k = F1 = 1.
(7) Phase Angle ( or Phase Lag)
From (5.4.8) we have
xp(t) =E1
√
(λ2 − ω2)2 + 4b2ω2
cos(ωt− θ)
= f(ω) · cos(ωt− θ) where
cos θ =λ2 − ω2
√
(λ2 − ω2)2 + 4b2ω2
sin θ =2bω
√
(λ2 − ω2)2 + 4b2ω2
The angle θ is called the phase angle or phase lag because it measures the lag of the output with
respect to the input. Fig. 5.10 shows the phase lag as a function of ω for the case m = k = 1. If
ω < ω0, then θ < π/2; if ω = ω0, then θ = π/2, and if ω > ω0, then θ > π/2.
tan θ =2bω
λ2 − ω2=
a/m · ωω20 − ω2
.
If m = k = 1 then ω0 = 1 and
θ = tan−1 aω
1− ω2.
Example
An 64-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The spring
constant is 18 lb/ft. The weight is then pulled down 6 in. below its equilibrium position and released
at t = 0 without any initial velocity. The external force is F (t) = 3 cosωt.
(1) Assuming the damping force is 4(dx/dt) please find the resonance frequency.
(2) Assuming there is no damping force please find the value of ω that give rise to undamped
resonance.
m = w/g = 64/32 = 2 (slug). k = 18 and F (t) = 3 cosωt.
2d2x
dt2+ a
dx
dt+ 18x = 3 cosωt
x(0) = 1/2, x′(0) = 0.
(1) a = 4
2d2x
dt2+ 4
dx
dt+ 18x = 3 cosωt
5.5 Resonance Phenomena 127
0 0.5 1 1.5 2 2.5 30
pi/2
pi
a=1.5
a=1.25
a=1.0
a=0.5
a=0.25
a=0
ω
θ (ω
)
Figure 5.10: The phase lag function of ω.
The resonance frequency is
1
2π
√
k
m− a2
2m2=
1
2π
√
18
2− 42
2 · 22
=1
2π
√7
.= 0.42 (cycles/sec)
Thus resonance occurs when ω =√7
.= 2.65.
(2) a = 0
2d2x
dt2+ 18x = 3 cosωt
Undamped resonance occurs when the frequency ω/2π of the impressed force is equal to the natural
frequency.
xc(t) = c1 sin 3t+ c2 cos 3t.
The natural frequency is 3/2π. Thus ω = 3 gives rise to undamped resonance. The ODE in this case
becomesd2x
dt2+ 9x = 3/2 cos 3t.
The initial conditions, x(0) = 1/2, x′(0) = 0, imply the solution
x(t) =1
3cos 3t+
1
4· t · sin 3t.
128 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
Example (The Details)
2d2x
dt2+ a
dx
dt+ 18x = 3 cosωt
(1) a = 4, x(0) = 0, x′(0) = 0, (2) a = 0, x(0) = 1/2, x′(0) = 0.
(1) a = 4
2m2 + 4m+ 18 = 0 m = (−2 ±√4− 4 · 9)/2 = −1±
√8
xc(t) = e−t(c1 sin√8t+ c2 cos
√8t).
λ2 = k/m = 18/2 = 9, 2b = a/2 = 2, b = 1 < λ = 3.
xc(t) = e−bt(c1 sin√
λ2 − b2t + c2 cos√
λ2 − b2t)
= e−t(c1 sin√8t + c2 cos
√8t)
x′c(t) = −e−t(c1 sin
√8t+ c2 cos
√8t) + e−t(
√8c1 cos
√8t−
√8c2 sin
√8t)
For the particular integral we have
xp(t) =E1
(λ2 − ω2)2 + 4b2ω2
[
(λ2 − ω2) cosωt+ 2bω sinωt]
=3/2
(9− ω2)2 + 4ω2
[
(9− ω2) cosωt+ 2ω sinωt]
x′p(t) =
3/2
(9− ω2)2 + 4ω2
[
2ω2 cosωt− (9− ω2)ω sinωt]
Solve the initial conditions and we obtain the solution
x(0) = 0 ⇒ c2 +3/2
(9− ω2)2 + 4ω2(9− ω2) = 0, c2 = −
3/2
(9− ω2)2 + 4ω2(9− ω2)
x′(0) = 0 ⇒ (−c2 +√8c1) +
3/2
(9− ω2)2 + 4ω2· 2ω2 = 0
c1 = − 3/2√8 [(9− ω2)2 + 4ω2]
(9 + ω2)
x(t) = xc(t) + xp(t)
= e−t
[
− 3/2 · (9 + ω2)√8 [(9− ω2)2 + 4ω2]
sin√8t− 3/2 · (9− ω2)
(9− ω2)2 + 4ω2cos√8t
]
+3/2
(9− ω2)2 + 4ω2
[
(9− ω2) cosωt+ 2ω sinωt]
(2) case 1. a = 0, ω 6= ω0 = λ =√
k/m = 3
2m2 + 0 ·m+ 18 = 0 m = ±3i
2d2x
dt2+ 18x = 3 cosωt
xc(t) = c1 sin 3t+ c2 cos 3t
5.5 Resonance Phenomena 129
λ2 = k/m = 18/2 = 9, 2b = a/2 = 0, b = 0 < λ = 3.
xc(t) = e−bt(c1 sin√
λ2 − b2t+ c2 cos√
λ2 − b2t)
= c1 sin 3t+ c2 cos 3t
x′c(t) = −3 · c1 sin 3t+ 3 · c2 cos 3t
For the particular integral we have
xp(t) =E1
(λ2 − ω2)2 + 4b2ω2
[
(λ2 − ω2) cosωt+ 2bω sinωt]
=E1
λ2 − ω2cosωt =
3/2
9− ω2· cosωt
x′p(t) = − 3/2
9 − ω2· ω sinωt
(a) Initial conditions: x(0) = 1/2, x′(0) = 0
x(0) = 1/2 ⇒ c2 +3/2
9− ω2= 1/2, c2 = 1/2− 3/2
9− ω2
x′(0) = 0 ⇒ 3c1 + 0 = 0, c1 = 0
x(t) = xc(t) + xp(t)
=
(
1
2− 3/2
9− ω2
)
cos 3t +3/2
9− ω2cosωt
(b) Initial conditions: x(0) = 0, x′(0) = 0
x(0) = 0 ⇒ c2 +3/2
9− ω2= 0, c2 = −
3/2
9− ω2
x′(0) = 0 ⇒ 3c1 + 0 + 0 = 0, c1 = 0
x(t) = xc(t) + xp(t)
= − 3/2
9− ω2cos 3t+
3/2
9− ω2cosωt
case 2. a = 0, ω = ω0 = λ =√
k/m = 3
2m2 + 0 ·m+ 18 = 0 m = ±3i
2d2x
dt2+ 18x = 3 cosωt
xc(t) = c1 sin 3t+ c2 cos 3t
130 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
UC set = t · sin 3t, t · cos 3t. For the particular integral we have
xp(t) =E1
(λ2 − ω2)2 + 4b2ω2
[
(λ2 − ω2) cosωt+ 2bω sinωt]
E1
λ2 − ω2cosωt
E1
2
√
m
k· t · sin
√
k
mt
=3/2
2
√
2
18· t · sin
√
18
2t =
1
4t sin 3t
x′p(t) =
1
4sin 3t+
3
4t cos 3t
x(t) = xc(t) + xp(t)
= c1 sin 3t+ c2 cos 3t+1
4t sin 3t
(a) Initial conditions: x(0) = 1/2, x′(0) = 0
x(0) = 1/2 ⇒ c2 + 0 = 1/2, c2 = 1/2
x′(0) = 0 ⇒ 3c1 + 0 = 0, c1 = 0
x(t) =1
2cos 3t +
1
4t sin 3t
(b) Initial conditions: x(0) = 0, x′(0) = 0
x(0) = 0 ⇒ c2 + 0 = 0, c2 = 0
x′(0) = 0 ⇒ 3c1 + 0 + 0 = 0, c1 = 0
x(t) =1
4t sin 3t
5.6 Electric Circuit Problems
1. Unifying power of mathematics
The mass-spring system has a strictly quantitative analogy in electrical engineering, namely, the
LRC-circuit. This will be an impressive demonstration of the unifying power of mathematics, illus-
trating that entirely different physical systems may have the same mathematical model and can be
treated and solved by the same method.
2. Electric circuit
The simplest electric circuit is a series circuit in which we have a source of electric energy
(electromotive force, EMF) such as a generator or a battery, and a resistor, which uses energy, for
5.6 Electric Circuit Problems 131
example, an electric light bulb.
SwitchE R
Fig. 5.10. A electrical network
If we close the switch, a current will flow through the resistor, and this will cause a voltage drop,
that is, electric potentials at the two ends of the resistor will be different.
3. Three important laws
The following three laws concerning the voltage drops across these various elements (resistor, induc-
tor, capacitor) are known to hold:
(1)
ER = R · i (5.6.1)
where R is a constant of proportionality called the resistance and i is the current.
(2)
EL = L · didt
(5.6.2)
where L is a constant of proportionality called the inductance and i is the current.
(3)
EC =1
C· q (5.6.3)
where C is a constant of proportionality called the capacitance and q is the instantaneous charge
on the capacitor. For the relations between q, i and EC we have
i =dq
dt
q =
∫
i dt
EC =1
C
∫
i dt
4. Conventional symbols
E: Electromotive force (battery or generator)
R: Resistor
L: Inductor
C: Capacitor
The units in common use are listed in Table 5.6.1.
Table 5.6.1 The units used in circuits
132 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
Quantity and symbol Unit
EMF or voltage E volt (V)
current i ampere (A)
charge q coulomb
resistance R ohm (Ω)
inductance L henry (H)
capacitance C farad (F)
5. Kirchhoff’s Laws
(1) Kirchhoff’s Voltage Law (KVL, Form 1 )
The algebraic sum of the instantaneous voltage drops around a closed circuit in a specific
direction is zero.
(2) Kirchhoff’s Voltage Law (KVL, Form 2 )
The algebraic sum of the voltage drops across resistors, inductors, and capacitors is equal to
the total EMF in a closed circuit.
(3) Kirchhoff’s Current Law (KCL)
At any point of a circuit, the sum of the inflowing currents is equal to the sum of the outflowing
currents.
6. Example (RL-circuit)
E
R
L
RL-circuit: Ldi
dt+R · i = E(t)
Case A. Constant EMF (direct current, DC)
5.6 Electric Circuit Problems 133
E(t) = E0 = Const. EL + ER = E(t)di
dt+
R
Li =
E0
Lµ(t) = exp
[∫
R
Ldt
]
= exp
(
R
Lt
)
µ · didt
+ µ · RLi = µ · E0
Ld(µ · i) = µ · E0
L
µ · i = E0
L
∫
µ dt+ c
i(t) =1
µ
[
E0
L
∫
µ dt+ c
]
= e−αt
[
E0
L
∫
eαt dt+ c
]
(α := R/L)
=E0
L· 1α+ e−αt · c =
E0
R+ c · e−Rt/L
The particular solution corresponding to the initial condition i(0) = 0 is
i(t) =E0
R(1− e−Rt/L)
=E0
R(1− e−t/τL)
where τL := L/R is called the inductive time constant of the circuit. See Fig. 5.11.
E0 / R
Time t
Cur
rent
i(t)
Figure 5.11: Current in an RL-circuit due to a constant EMF .
Case B. Periodic EMF (alternating current, AC)
Method I. Linear equation and integrating factor
134 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
We consider the integrating factor eαt where αdef= R/L.
E(t) = E0 sinωt
i(t) = e−αt
[
E0
L
∫
eαt sinωt dt+ c
]
(α = R/L)
= ce−Rt/L +E0
R2 + ω2L2(R sinωt− ωL cosωt)
= ce−Rt/L +E0√
R2 + ω2L2sin(ωt− δ)
where δ = tan−1(ωL/R).
Check
e−αt
[
E0
L
∫
eαt sinωt dt+ c
]
(α = R/L)
= ce−Rt/L +E0
R2 + ω2L2(R sinωt− ωL cosωt)
Integration by parts. u = sinωt, dv = eαt dt, du = ω cosωt dt∫
eαt sinωt dt =1
αsinωt · eαt − ω
α
∫
eαt cosωt dt (u = cosωt, dv = eαt dt)
=1
αsinωt · eαt − ω
α
[
1
αcosωt · eαt + ω
α
∫
eαt sinωt dt
]
Therefore(
1 +ω2
α2
)∫
eαt sinωt dt =1
αsinωt · eαt − ω
α2cosωt · eαt
∫
eαt sinωt dt =α2
α2 + ω2
[
1
αsinωt · eαt − ω
α2cosωt · eαt
]
=1
α2 + ω2
[
α sinωt · eαt − ω cosωt · eαt]
Thus, we have
e−αt
[
E0
L
∫
eαt sinωt dt+ c
]
= ce−αt + e−αt · E0
L· 1
α2 + ω2
[
α sinωt · eαt − ω cosωt · eαt]
= ce−Rt/L +E0
L· 1
(R/L)2 + ω2
[
R
Lsinωt− ω cosωt
]
= ce−Rt/L +E0
R2 + ω2L2(R sinωt− ωL cosωt)
Method II. UC method
5.6 Electric Circuit Problems 135
Let ip(t) = A cosωt+B sinωt. Then i′p(t) = Bω cosωt−Aω sinωt. Substituting them to the original
equation we have
LBω + RA = 0
−LAω + RB = E0
⇒
(R) ·A + (Lω) ·B = 0
(−Lω) · A + (R) · B = E0
By Cramer’s rule we have
=
∣
∣
∣
∣
∣
R Lω
−Lω R
∣
∣
∣
∣
∣
= R2 + ω2L2 A =
∣
∣
∣
∣
∣
0 Lω
E0 R
∣
∣
∣
∣
∣
B =
∣
∣
∣
∣
∣
R 0
−Lω E0
∣
∣
∣
∣
∣
Thus A =−ωLE0
R2 + ω2L2, B =
E0R
R2 + ω2L2and
ip(t) =E0
R2 + ω2L2(R sinωt− ωL cosωt)
For the complementary function we know the characteristic equation is Lλ+R = 0. So λ = −R/L.
ic(t) = c · e−RL
Thus we have the solution
i(t) = ic(t) + ip(t) = ce−Rt/L +E0
R2 + ω2L2(R sinωt− ωL cosωt).
Interpretation. The exponential term will approach zero as t approaches to infinity. This means
that after sufficiently long time the current i(t) executes practically harmonic oscillations. The
oscillations of i(t) are in phase with those of EMF E(t) with a phase angle lag δ. See Fig. 5.12. The
phase angle δ as a function of ωL/R. If L = 0 then δ = 0. The oscillations of i(t) are in phase with
those of EMF E(t). See Fig. 5.13. Note that
(1) An electrical (or dynamical) system is said to be in the steady state when the variables describing
its behavior are periodic functions of time or constant, and it is said to be in the transient state
(or unsteady state) when it is not in the steady state. The corresponding variables are called
steady-state functions and transient functions, respectively.
(2) For the case A, E0/R is a steady-state function (or a steady-state solution). For the case B,
E0
R2 + ω2L2sin(ωt− δ)
is a steady-state function. Before the circuit reaches the steady state it is in the transient state.
(3) Such an interim or transient period occurs because inductors and capacitors store energy, and the
corresponding inductor currents and capacitor voltages can not be changed instantly. Practically,
this transient state will last only a short time.
7. Example
A circuit has in series an electromotive force given by E = 100 sin 40t V, a resistor of 10 Ω and an
136 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
0 \pi 2 \pi 3 \pi 4 \p 5 \pi−1
−0.5
0
0.5
1
1.5
2
Current i(t)
Exponential term
Time t
Cur
rent
i(t)
Figure 5.12: Current in an RL-circuit due to a periodic EMF (δ = π/4).
inductor of 0.5 H. If the initial current is 0, find the current at time t > 0.
0.5di
dt+ 10i = 100 sin 40t
di
dt+ 20i = 200 sin 40t
µ(t) = exp[
∫
20 dt] = e20t
i(t) = 2(sin 40t− 2 cos 40t) + c · e−20t
i(0) = 0 ⇒ c = 4
i(t) = 2(sin 40t− 2 cos 40t) + 4 · e−20t
Expressing the trigonometric terms in a ”phase-angle” form, we have
i(t) = 2√5
(
1√5sin 40t− 2√
5cos 40t
)
+ 4 · e−20t
i(t) = 2√5 sin(40t+ φ) + 4 · e−20t,
where φ is determined by
φ = cos−1(1/√5) = sin−1(−2/
√5)
.= −1.11 rad
i(t) = 2√5 sin(40t− 1.11) + 4 · e−20t.
5.6 Electric Circuit Problems 137
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
π /2
ω L/ R
δ
δ = arctan( ω L/ R )
Figure 5.13: The phase angle function δ = tan−1(ωL/R).
Note that
(1) the phase angle φ.= −1.11 rad.
(2) the period π/20 is the same as that of E(t).
(3) the E(t) leads the steady-state current by approximately 1.11/40.
(sin(40t− 1.11) = sin[40(t− 1.11/40)])
8. LRC-circuit
We now consider the circuit shown in Fig. 5.14.
E
R
L
C Fig. 5.14 LRC-circuit
The ordinary differential equation is
Ldi
dt+Ri+
1
Cq = E (5.6.4)
Ld2q
dt2+R
dq
dt+
1
Cq = E. (i =
dq
dt) (5.6.5)
138 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
On the other hand, if we differentiate Eqn. (5.6.4) w.r.t. t, then we have
Ld2i
dt2+R
di
dt+
1
C
dq
dt=
dE
dt(5.6.6)
Ld2i
dt2+R
di
dt+
1
Ci =
dE
dt. (i =
dq
dt) (5.6.7)
Note that
(1) RL-circuit: if the circuit contains no capacitor, Eqn. (5.6.4) reduces to
Ldi
dt+Ri = E. (5.6.8)
(2) RC-circuit: if the circuit contains no inductor, Eqn(5.6.4) reduces to
Rdq
dt+
1
Cq = E. (5.6.9)
9. Unifying power of mathematics
The important comparison of the mechanical system and the electric system
An interesting and useful analogy for these two systems
Ld2q
dt2+R
dq
dt+
1
Cq = E(t) (5.6.10)
md2x
dt2+ a
dx
dt+ kx = F (t). (5.6.11)
See Table 5.6.2.
Table 5.6.2 The comparison of the mechanical system and the electric system
Mechanical system Electrical system
mass m inductance L
damping constant a resistance R
spring constant k reciprocal of capacitance 1/C
impressed force F (t) impressed voltage or EMF E(t)
displacement x charge q
velocitydx
dt= v current
dq
dt= i
10. Example
A circuit has in series an electromotive force given by E = 100 sin(60t) V, a resistor of 2Ω, an
inductor of 0.1 H, and a capacitor of 1/260 farads. (See Fig. 5.14) If the initial current and the
initial charge on the capacitor are both zero, find the charge on the capacitor at any time t > 0.
5.6 Electric Circuit Problems 139
Answer. (1) Formulation L = 0.1, R = 2, C = 1/260, E = 100 sin(60t)
0.1d2q
dt2+ 2
dq
dt+ 260q = 100 sin(60t)
d2q
dt2+ 20
dq
dt+ 2600q = 1000 sin(60t)
q(0) = 0 , q′(0) = 0.
(2) Solution The auxiliary equation is
r2 + 20r + 2600 = 0, r = −10 ± 50i
The complementary function is
qc = e−10t(c1 sin 50t+ c2 cos 50t).
Employing the method of undetermined coefficients to find a particular integral, we write
qp = A sin 60t+B cos 60t.
Differentiating twice and substituting into the ODE we find that
A = −2561
and B = −3061
The general solution is
q(t) = qc(t) + qp(t)
= e−10t(c1 sin 50t+ c2 cos 50t) +
(
−2561
)
sin 60t+
(
−3061
)
cos 60t.
q(0) = 0 and q′(0) = 0 imply c1 =36
61and c2 =
30
61,
q(t) = e−10t
(
36
61sin 50t+
30
61cos 50t
)
+
(
−2561
)
sin 60t+
(
−3061
)
cos 60t
=6√61
61e−10t cos(50t− φ)− 5
√61
61cos(60t− θ)
.= 0.77e−10t cos(50t− 0.88)− 0.64 cos(60t− 0.69).
(3) Interpretation
The first term is the transient term. The second term is the steady-state term.
11. Reactance and Impedance
Consider the solution of the ordinary differential equation
Ldi
dt+Ri+
1
Cq = E. (5.6.12)
140 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
For a sinusoidal EMF E(t) = E0 sinωt we have dE/dt = E0ω cosωt.
ip(t) = a cosωt+ b sinωt. (5.6.13)
Substituting this into (5.6.12) we obtain
a =−E0S
R2 + S2, b =
E0R
R2 + S2(5.6.14)
where S is the so-called reactance given by
S = ωL− 1
ωC. (5.6.15)
check!
In any practical case, R 6= 0 so that the denominators of a and b are not zeros. Moreover, ip(t) =
i0 sin(ωt− θ) where
i0 =√a2 + b2 =
E0√R2 + S2
, tan θ = −ab=
S
R. (5.6.16)
The quantity√R2 + S2 is called the impedance which is somewhat analogous to the resistance
R = E/i (Ohm’s law) because√R2 + S2 =
E0
i0. (5.6.17)
The complementary function is
ic = c1eλ1t + c2e
λ2t
where λ1 and λ2 are the roos of the characteristic equation λ2 +R/Lλ+1/LC = 0. So λ1 = −α+ β
and λ2 = −α− β where
α =R
2L, β =
1
2L
√
R2 − 4L
C. (5.6.18)
If R > 0, ic(t)→ 0 as t→∞.
Hence, the transient current i = ic+ ip tends to steady-state current ip, and after some tie the output
will practically be a harmonic oscillation, which is given by ip and whose frequency is that of the
input.
12. Example
Find the current i(t) in an LRC-circuit with R = 100Ω, L = 0.1H , C = 10−3 Farads, E(t) =
155 sin 377t (hence 377/2π = 60Hz = 60 cycles/sec) and zero charge and current when time t = 0.
Find the reactance and the general solution.
Answer.
0.1d2i
dt2+ 100
di
dt+ 1000i = 155 · 377 · cos 377t.
5.6 Electric Circuit Problems 141
The reactance S = 37.7− 1/0.377 = 35.
The steady-state current ip(t) = a cos 377t+ b sin 377t.
a =−155 · 351002 + 352
= −0.483, b =−155 · 1001002 + 352
= 1.381.
We have ip(t) = −0.483 cos 377t+ 1.381 sin 377t = 1.463 sin(377t− 0.34).
The characteristic equation is 0.1λ2 + 100λ+ 1000 = 0. Thus, λ1 = −10 and λ2 = −990.The general solution is
i(t) = c1e−10t + c2e
−990t + 1.463 sin(377t− 0.34).
The initial conditions q(0) = 0 and i(0) = 0 imply c1 = −0.043 and c2 = 0.526. Thus,
i(t) = −0.043e−10t + 0.526e−990t + 1.463 sin(377t− 0.34).
13. A Series Circuit and A Parallel Circuit
For a series LRC-circuit we know that the differential equation is
Ld2i
dt2+R
di
dt+
1
C
dq
dt=
dE
dt, (5.6.6)
Ld2i
dt2+R
di
dt+
1
Ci =
dE
dt. (i =
dq
dt) (5.6.7)
E
R
L
C Fig. 5.14 A series LRC-circuit
For a parallel LRC-circuit we know that applying Kirchhoff’s current law to the network in Figure
5.15 leads us to the equation
IS(t) =1
RV + C
dV
dt+
1
L
∫
V (s)ds. (5.6.19)
Upon differentiating and rearranging terms, the equation becomes
d2V
dt2+
1
RC
dV
dt+
1
LCV =
1
C
dIS(t)
dt. (5.6.20)
IS(t)Figure 5.15. A parallel
LRC-electrical network .
142 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
14. Example
Use the following consistent set of scaled units (referred to as the Scaled SI Unit System).
Table 5.6.3 The scaled units
Quantity Unit Symbol
Voltage volt V
Current milliampere (mA) I
Time millisecond (ms) t
Resistance kilohm (kΩ) R
Inductance henry (H) L
Capacitance microfarad (µF) C
Use the consistent set of scaled units as above. Consider the parallel LRC network shown in Fig.
5.16. Assume that at time t = 0, the voltage V (t) and its time rate of change are both zero. For the
given source voltage IS(t) = 1− e−t mA, determine the voltage V (t).
IS(t)C = 0.5µFR= 1 kΩL= 1 H
Figure 5.16. The parallel
LRC-electrical network .
IS(t) =1
RV + C
dV
dt+
1
L
∫
V (s)ds.
Upon differentiating and rearranging terms, the equation becomes
d2V
dt2+
1
RC
dV
dt+
1
LCV =
1
C
dIS(t)
dt,
V ′′ + 2V ′ + 2V = 2e−t.
Vc(t): λ2 + 2λ+ 2 = 0, λ = −1± i.
Vc(t) = c1e−t cos t + c2e
−t sin t.
Vp(t): Let Vp(t) = Ae−t. A− 2A+ 2A = 2, A = 2, Vp(t) = 2e−t.
V (t) = Vc(t) + Vp(t) = c1e−t cos t + c2e
−t sin t+ 2e−t.
5.6 Electric Circuit Problems 143
Applying initial conditions we obtain
V (0) = 0 = c1 + 2
V ′(0) = 0 = −c1 + c2 − 2.
Thus c1 = −2, c2 = 0. The solution V (t) is then
V (t) = −2e−t cos t+ 2e−t,
= 2e−t(1− cos t). (Voltage)
144 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE
Chapter 6
Systems of Linear Differential Equations
6.1 Basic Theory of Linear Systems
A. Introduction
1. Normal Form
The normal form in the case of two first-order ODEs in two unknown functions x and y is
dx
dt= a11(t)x+ a12(t)y + F1(t), (6.1.1)
dy
dt= a21(t)x+ a22(t)y + F2(t).
If F1(t) and F2(t) are zero for all t, then the system is called homogeneous; otherwise, the
system is said to be nonhomogeneous.
2. Example 6.1.1 (a)
dx
dt= 2x− y, (6.1.2)
dy
dt= 3x+ 6y,
is homogeneous.
Example 6.1.1 (b)
dx
dt= 2x− y − 5t, (6.1.3)
dy
dt= 3x+ 6y − 4,
is nonhomogeneous.
146 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
3. Definition 6.1.1
By a solution of system (6.1.1) we shall mean an ordered pair (f, g) such that
df(t)
dt= a11(t)f(t) + a12(t)g(t) + F1(t),
dg(t)
dt= a21(t)f(t) + a22(t)g(t) + F2(t).
for all t such that a ≤ t ≤ b.
4. Example 6.1.2
The ordered pair (e5t,−3e5t) is a solution of the system (6.1.2). Note that (e3t,−e3t) is also a
solution of the system (6.1.2).
5. Theorem 6.1.1
Hypothesis. Let a11, a12, F1, a11, a11, F2 be continuous on a ≤ t ≤ b, t0 ∈ [a, b] and c1 and
c2 be two constants.
Conclusion. There exists a unique solution (x = f(t), y = g(t)) of system (6.1.1) such that
f(t0) = c1 and g(t0) = c2 and this solution is defined on the entire interval a ≤ t ≤ b.
B. Homogeneous Linear System
1. Consider the following homogeneous linear system
dx
dt= a11(t)x+ a12(t)y, (6.1.4)
dy
dt= a21(t)x+ a22(t)y.
Theorem 6.1.2
Hypothesis. Let (x = f1(t), y = g1(t)) and (x = f2(t), y = g2(t)) be two solutions, and c1 and
c2 be two constants.
Conclusion. Then
x = c1f1(t) + c2f2(t), (6.1.5)
y = c1g1(t) + c2g2(t),
is also a solution of the system (6.1.4).
2. Definition 6.1.2 (Linear Combination)
The solution (6.1.5) is called a linear combination of the solutions (f1(t), g1(t)) and (f2(t), g2(t)).
3. Theorem 6.1.3 (Restated)
Any linear combination of two solutions of the homogeneous linear system (6.1.4) is itself a
solution of the system (6.1.4).
6.1 Basic Theory of Linear Systems 147
4. Example 6.1.3
(e5t,−3e5t) and (e3t,−e3t) are two solutions of the system (6.1.2). Then
x = c1e5t + c2e
3t,
y = −3c1e5t − c2e3t,
is also a solution of the system (6.1.2).
5. Definition 6.1.3 (Linear Dependence)
Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solution of the system (6.1.4). These two solutions
are linearly dependent on t ∈ [a, b] if ∃ c1, c2 not both zero, such that
c1f1(t) + c2f2(t) = 0,
c1g1(t) + c2g2(t) = 0,
for all a ≤ t ≤ b.
6. Definition 6.1.4 (Linear Independence)
Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solutions of the system (6.1.4). These two solution
are linearly independent on t ∈ [a, b] if
c1f1(t) + c2f2(t) = 0,
c1g1(t) + c2g2(t) = 0,
implies c1 = c2 = 0.
7. Example 6.1.4 (a)
(e5t,−3e5t) and (2e5t,−6e3t) are linearly dependent.
Example 6.1.4 (b)
(e5t,−3e5t) and (e3t,−e3t) are linearly independent.
8. Theorem 6.1.4
There exists sets of two linearly independent solutions of the homogeneous linear system (6.1.4).
Every solution of the system (6.1.4) can be written as a linear combination of any two linearly
independent solution if (6.1.4).
9. Definition 6.1.5 (General Solution)
Let (f1(t), g1(t)) and (f2(t), g2(t)) be two linearly independent solutions of the system (6.1.4).
Then
x = c1f1(t) + c2f2(t),
y = c1g1(t) + c2g2(t),
is called the general solution of the system (6.1.4).
148 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
10. Theorem 6.1.5
Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solutions of the system (6.1.4). Then a necessary and
sufficient condition that two solutions be linear independent on t ∈ [a, b] is that the determinant
(t) =
∣
∣
∣
∣
∣
f1(t) f2(t)
g1(t) g2(t)
∣
∣
∣
∣
∣
6= 0,
for all t ∈ [a, b].
11. Theorem 6.1.6
The determinant (t) of Theorem 6.1.5 either is identically zero or vanishes for no value of t
on the interval a ≤ t ≤ b.
12. Example 6.1.5 (a)
Consider the two solutions, (e5t,−3e5t) and (2e5t,−6e3t). (t) = 0 for all t ∈ [a, b].
Example 6.1.5 (b)
Consider the two solutions,(e5t,−3e5t) and (e3t,−e3t).
(t) =
∣
∣
∣
∣
∣
e5t e3t
−3e5t −e3t
∣
∣
∣
∣
∣
= 2e8t 6= 0,
for all t ∈ [a, b].
C. Nonhomogeneous Linear System
1. Consider the following nonhomogeneous linear system
dx
dt= a11(t)x+ a12(t)y + F1(t), (6.1.1)
dy
dt= a21(t)x+ a22(t)y + F2(t).
Theorem 6.1.7
Hypothesis. Let (x = f0(t), y = g0(t)) be any solution of the system (6.1.1) , and (x =
f(t), y = g(t)) be any solution of the corresponding homogeneous system (6.1.4).
Conclusion. Then
x = f1(t) + f0(t),
y = g1(t) + g0(t),
is also a solution of the system (6.1.1).
6.1 Basic Theory of Linear Systems 149
2. Definition 6.1.6 (General Solution)
Let (x = f0(t), y = g0(t)) be any solution of the nonhomogeneous system (6.1.1) , and
(x = f1(t), y = g1(t)) and (x = f2(t), y = g2(t)) be two linearly independent solutions of
the corresponding homogeneous system (6.1.4). Then
x = c1f1(t) + c2f2(t) + f0(t), (6.1.6)
y = c1g1(t) + c2g2(t) + g0(t),
is called the general solution of the system (6.1.1).
3. Example 6.1.6
dx
dt= 2x− y − 5t, (6.1.3)
dy
dt= 3x+ 6y − 4,
Sol. (2t+ 1,−t) is a solution of the nonhomogeneous system.
(e5t,−3e5t) and (e3t,−e3t) are two linearly independent solutions of the corresponding homo-
geneous system. Then the general solution is
x = c1e5t + c2e
3t + 2t+ 1
y = −3c1e5t − c2e3t − t
D. A Solution Method for Homogeneous Linear Systems with Constant Coefficients
1. Consider the following nonhomogeneous linear system
dx
dt= a1x+ b1y, (6.1.7)
dy
dt= a2x+ b2y.
Let us attempt to determine a solution of the form
x = Aeλt,
y = Beλt.
150 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
Then we have
dx
dt= Aλeλt = a1Ae
λt + b1Beλt,
dy
dt= Bλeλt = a2Ae
λt + b2Beλt.
We need to solve
(a1 − λ)A + b1B = 0,
a2A+ (b2 − λ)B = 0.
This system has a nontrivial solution if
(t) =
∣
∣
∣
∣
∣
a1 − λ b1
a2 b2 − λ
∣
∣
∣
∣
∣
= 0.
Then
pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0. (6.1.8)
This equation is called the characteristic equation associated with the system (6.1.4). In matrix
form,
d−xdt
def=
(
dxdtdydt
)
=
(
a1 b1
a2 b2
)(
x
y
)
def= M−x (6.1.4)
We want to determine a solution of the form
−x =
(
A
B
)
eλt.
Then we have
d−xdt
=
(
A
B
)
λ · eλt =(
a1 b1
a2 b2
)(
x
y
)
= M−x = M
(
A
B
)
eλt.
That is,
M
(
A
B
)
= λ ·(
A
B
)
.
We must find the eigenvalues, λ, and the corresponding eigenvectors, (A,B), of the matrix M
by the characteristic equation
pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0. (6.1.8)
2. Example 6.1.7
dx
dt= 2x− y, (6.1.2)
dy
dt= 3x+ 6y,
6.2 Matrices and Vectors 151
pM(λ) = λ2 − (2 + 6)λ+ (2 · 6− (−1) · 3) = 0, λ = 3, 5.
λ = 3:
A− 3I =
(
2− 3 −13 6− 3
)
=
(
−1 −13 3
)
.
The eigenvector is (1,−1).λ = 5:
A− 5I =
(
2− 5 −13 6− 5
)
=
(
−3 −13 1
)
.
The eigenvector is (1,−3).The general solution is
−x = c1
(
1
−1
)
e3t + c2
(
1
−3
)
e5t.
6.2 Matrices and Vectors
A. Matrix Multiplication and Inversion
1. Definition 6.2.1 (Inverse)
Let A be an n× n nonsingular matrix. The unique n× nmatrix B satisfies
AB = BA = In,
is called the inverse of A. We use the notation:
AA−1 = A−1A = In.
2. Definition 6.2.2 (Minor, Cofactor and the Matrix of Cofactor)
Let A ∈Mn(F) (i.e., an n× n matrix). A = (aij)n×n
(1) The minor of aij is defined to be the determinant of the (n− 1)× (n− 1) matrix obtained
from A by deleting the ith row and jth column of A. We denote it by mij .
(2) The cofactor of aij is defined to be the minor Mij of aij multiplied by the number (−1)i+j .
We denote it by cij .
cij = (−1)i+jmij . (6.2.1)
(3) The matrix of cofactor of the elements of A is defined to be the matrix obtained from A by
replacing each element aij of A by its cofactor cij. We denote it by cof A.
152 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
3. Example 6.2.1
A =
2 1 −14 3 −2−6 2 5
. Find cof A.
m23 =
∣
∣
∣
∣
∣
2 1
−6 2
∣
∣
∣
∣
∣
= 10. c23 = (−1)2+3m23 = −10.
Thus, the matrix of cofactor is
cof A =
∣
∣
∣
∣
∣
3 −22 5
∣
∣
∣
∣
∣
−∣
∣
∣
∣
∣
4 −2−6 5
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
4 3
−6 2
∣
∣
∣
∣
∣
−∣
∣
∣
∣
∣
1 −12 5
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2 −1−6 5
∣
∣
∣
∣
∣
−∣
∣
∣
∣
∣
2 1
−6 2
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 −13 −2
∣
∣
∣
∣
∣
−∣
∣
∣
∣
∣
2 −14 −2
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2 1
4 3
∣
∣
∣
∣
∣
=
19 −8 26
−7 4 −101 0 2
.
4. Definition 6.2.3 (Classical Adjoint)
Let A ∈ Mn(F). The classical adjoint of A is defined to be the transport of cof A. We denote
it by adj A (or A@).
5. Example 6.2.2
A =
2 1 −14 3 −2−6 2 5
. Find adj A.
adj A = (cof A)T =
19 −8 26
−7 4 −101 0 2
T
=
19 −7 1
−8 4 0
26 −10 2
.
6. Result A
Let A be an n× n nonsingular matrix. Then
A−1 =1
det(A)( adj A ),
where det(A) is the determinant of A and adj A is the classical adjoint matrix of A.
6.2 Matrices and Vectors 153
7. Example 6.2.3
A =
2 1 −14 3 −2−6 2 5
. Find A−1.
A−1 =1
det(A)( adj A ) =
1
4
19 −8 26
−7 4 −101 0 2
T
=1
4
19 −7 1
−8 4 0
26 −10 2
.
8. Definition 6.2.4 (Eigenvalues and Eigenvectors)
An eigenvalue (or characteristic value) of the matrix A is a number λ for which the equation
A−v = λ−v
has a nonzero solution −v .
An eigenvector (or characteristic vector) of the matrix A is a nonzero −v such that A−v = λ−vfor some number λ.
9. Example 6.2.4
A =
(
6 −32 1
)
, λ = 4, −v =
(
3
2
)
.
A−v =
(
6 −32 1
)(
3
2
)
=
(
12
8
)
= 4 ·(
3
2
)
= λ · −v
Thus λ = 4 is an eigenvalue and −v = (3, 2) 6= −0 is a corresponding eigenvector.
10. Result B
Solve the problem of determining the eigenvalues and eigenvectors.
A−v = λ−v has nonzero solutions −v if and only if (A− λI)−v = 0 has nonzero solutions, i.e.,
det(A− λI) = 0.
Definition 6.2.5
The characteristic polynomial of A is
pA(λ) = det(A− λI)
in the variable λ. The characteristic equation of A is pA(λ) = det(A− λI) in the unknown λ.
The eigenvalues of A are the roots.
154 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
11. Example 6.2.5
A =
(
6 −32 1
)
, Find all eigenvalues and eigenvectors.
det(A− λI) = 0, pA(λ) = λ2 − 7λ+ 12 = 0, λ = 4, 3.
λ1 = 4, (A− 4I)−v1 =
(
2 −32 −3
)(
x1
x2
)
=
(
0
0
)
We choose −v1 =
(
3
2
)
.
λ2 = 3, (A− 3I)−v2 =
(
3 −32 −2
)(
x1
x2
)
=
(
0
0
)
We choose −v2 =
(
1
1
)
.
Thus λ = 4 is an eigenvalue and −v = (3, 2) 6= −0 is a corresponding eigenvector and λ = 3 is
another eigenvalue and −v = (1, 1) 6= −0 is a corresponding eigenvector.
Let B1 = v1, v2 be a basis of R2,
P1 =
(
3 1
2 1
)
, P−11 =
(
1 −1−2 3
)
, and P−11 AP1 =
(
4 0
0 3
)
.
Let B2 = v2, v1 be a basis of R2,
P2 =
(
1 3
1 2
)
, P−12 =
1
(−1)
(
2 −3−1 1
)
, and P−12 AP2 =
(
3 0
0 4
)
.
The matrix A can be diagonalized.
12. Result C
Suppose each of the n eigenvalues λ1, . . . , λn of the matrix A is distinct (i.e., non-repeated);
and let v1, v2, . . . , vn be a set of respective corresponding eigenvectors of A. Then
v1, v2, . . . , vn is linearly independent.
13. Result D
Suppose that A ∈ Mn(F) has a eigenvalue of multiplicity m, where 1 < m ≤ n. Then
this repeated eigenvalue having multiplicity m has p linearly independent eigenvectors corre-
sponding to it, where
1 ≤ p ≤ m.
14. Result E
If A ∈Mn(F) has one or more repeated eigenvalues, then
there may exist less than n linearly independent eigenvectors of A.
6.2 Matrices and Vectors 155
15. Example 6.2.6
A =
1 1 2
0 1 3
0 0 2
, Find all eigenvalues and eigenvectors.
det(A− λI) = 0, pA(λ) = (λ− 1)2(λ− 2) = 0, λ = 1, 1, 2.
λ1 = 1, (A− I)−v1 =
0 1 2
0 0 3
0 0 2
x1
x2
x3
=
0
0
0
We choose −v1 =
1
0
0
.
λ2 = 2, (A− 2I)−v2 =
−1 1 2
0 −1 3
0 0 0
x1
x2
x3
=
0
0
0
We choose −v2 =
5
3
1
.
There exists less than 3 linearly independent eigenvectors of A. Note that A can not be
diagonalized.
16. Result F
All of the eigenvalues of a real symmetric matrix are real numbers.
17. Result G
If A ∈Mn(F) is symmetric, then
there exist n linearly independent eigenvectors of A, whether the n eigenvalues of A are all
distinct or whether one or more of these eigenvalues is repeated.
18. Example 6.2.7
A =
7 −1 6
−10 4 −12−2 1 −1
, Find all eigenvalues and eigenvectors.
156 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
det(A− λI) = 0, pA(λ) = (λ− 2)(λ− 3)(λ− 5) = 0, λ = 2, 3, 5.
λ1 = 2, (A− 2I)−v1 =
5 −1 6
−10 2 −12−2 1 −3
x1
x2
x3
=
0
0
0
We choose −v1 =
1
−1−1
.
λ2 = 3, (A− 3I)−v2 =
4 −1 6
−10 1 −12−2 1 −4
x1
x2
x3
=
0
0
0
We choose −v2 =
1
−2−1
.
λ3 = 5, (A− 5I)−v3 =
2 −1 6
−10 −1 −12−2 1 −6
x1
x2
x3
=
0
0
0
We choose −v3 =
3
−6−2
.
There exists 3 linearly independent eigenvectors of A. Note that A can be diagonalized.
19. Result H
Let Sk ∈Mk(F).
Sk =
0
1 0
1 0. . .
. . .
1 0
, STk =
0 1
0 1. . .
. . .
0 1
0
.
They are called the elementary nilpotent matrices. We note that
Sk
x1
x2
...
xk
=
0
x1
...
xk−1
, STk
x1
x2
...
xk
=
x2
...
xk
0
.
Let A ∈Mn(F). If B1 = v1, v2, . . . , vn satisfies that
(A− λI)n(vn) = 0, (A− λI)n−1(vn) 6= 0, and
vn−1 = (A− λI)vn
vn−2 = (A− λI)vn−1
· · ·v1 = (A− λI)v2
then (1) B1 is a (cyclic) basis for Fn.
(2) [A]B1 = STn + λIn (a Jordan fundamental matrix).
(3) Let B2 = vn, vn−1, . . . , v1. Then B2 is also a (cyclic) basis for Fn and [A]B2 = Sn+λIn.
6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns 157
20. Example 6.2.8
A =
0 1 2
−5 −3 −71 0 0
, Find the Jordan form.
det(A− λI) = 0, pA(λ) = −(λ+ 1)3 = 0, λ = −1, −1, −1.
(A + I) =
1 1 2
−5 −2 −71 0 1
, (A+ I)2 =
−2 −1 −3−2 −1 −32 1 3
, (A+ I)3 = 0.
We should choose v3 = (1, 0, 0) ∈ ker(A+ I)3 − ker(A+ I)2. Then v2 = (A+ I)v3 = (1,−5, 1)and v1 = (A+ I)v2 = (−2,−2, 2).B1 = v1, v2, v3,
P1 =
−2 1 1
−2 −5 0
2 1 0
, P−1
1 =
0 1/8 5/8
0 −1/4 −1/41 1/2 3/2
, and P−1
1 AP1 =
−1 1 0
0 −1 1
0 0 −1
.
B2 = v3, v2, v1,
P2 =
1 1 −20 −5 −20 1 2
, P−1
2 =
1 1/2 3/2
0 −1/4 −1/40 1/8 5/8
, and P−1
2 AP2 =
−1 0 0
1 −1 0
0 1 −1
.
6.3 Homogeneous Linear Systems with Constant Coefficients:
Two Unknowns
Let us consider the homogeneous linear system with constant coefficients:
dx
dt= a1x+ b1y, (6.3.1)
dy
dt= a2x+ b2y.
We attempt to obtain a solution of the form
−x =
(
x(t)
y(t)
)
=
(
A
B
)
eλt
158 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
It need to consider the characteristic equation (6.1.8)
pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0.
A. Case 1. The Roots of the Characteristic Equation are Real and Distinct
1. Theorem 6.3.1
Hypothesis. The roots λ1 and λ2 are real and distinct. −v1 = (A1, B1) and−v2 = (A2, B2) are
corresponding eigenvectors with respect to λ1 and λ2
Conclusion. The system (6.3.1) has two nontrivial linearly independent solutions. The general
solution is
−x =
(
x(t)
y(t)
)
= c1eλ1t−v1+c2e
λ2t−v2 = c1
(
A1
B1
)
eλ1t+c2
(
A2
B2
)
eλ2t =
(
c1A1eλ1t + c2A2e
λ2t
c1B1eλ1t + c2B2e
λ2t
)
.
2. Example 6.3.1
dx
dt= 6x− 3y, (6.3.2)
dy
dt= 2x+ y,
pA(λ) = λ2 − (6 + 1)λ+ (1 · 6− (−3) · 2) = 0, λ = 3, 4.
λ = 3:
A− 3I =
(
6− 3 −32 1− 3
)
=
(
3 −32 −2
)
.
The eigenvector is (1, 1).
λ = 4:
A− 4I =
(
6− 4 −32 1− 4
)
=
(
2 −32 −3
)
.
The eigenvector is (3, 2).
The general solution is
−x = c1
(
1
1
)
e3t + c2
(
3
2
)
e4t.
B. Case 2. The Roots of the Characteristic Equation are Real and Equal
6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns 159
1. Theorem 6.3.2
Hypothesis. The roots λ1 and λ2 are real and equal. Let λ denote their common value.
Further assume that the system (6.3.1) is not such that a1 = b2 6= 0 and a2 = b1 = 0.
Conclusion. The system (6.3.1) has two nontrivial linearly independent solutions of the form
−x1 =
(
x1(t)
y1(t)
)
=
(
A
B
)
eλt and −x2 =
(
x2(t)
y2(t)
)
=
(
A1t + A2
B1t +B2
)
eλt
where A1 and B1 are not both zero and B1/A1 = B/A.
The general solution is
−x =
(
x(t)
y(t)
)
= c1
(
A
B
)
eλt + c2
(
A1t + A2
B1t +B2
)
eλt.
Method II (A Cyclic Basis)
We can choose −v1 and −v2 such that
(A− λI)2 −v2 =−0 ,
(A− λI) −v2 = −v1 ,
and −v1 and −v2 are linearly independent. Then
the two solutions
−x1 = −v1 eλt,
−x2 = (−v1 t +−v2 )eλt,
are linearly independent.
Method III (Another Notations)
We can choose−ξ and −η such that
(A− λI)−ξ =
−0 ,
(A− λI) −η =−ξ ,
Then
the two solutions
−x1 =−ξ eλt,
−x2 = (−ξ t +−η )eλt,
are linearly independent.
Remark:−ξ ≡ −v1 and −η ≡ −v2 .
160 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
Proof. Let −x = c1−x1 + c2
−x2 = c1−ξ eλt + c2(
−ξ t +−η )eλt.
Then we check the first equation. (Note that A−ξ = λ
−ξ .)
d −x1
d t= (−ξ · λ) · eλt
= (A−ξ ) · eλt = A−x1.
Now we check the second equation. (Note that A−η = λ−η +−ξ .)
d −x2
d t= (−ξ ) · eλt + (
−ξ t+−η )λ · eλt
= (−ξ +−ξ t · λ+−η · λ)eλt
= (A−ξ · t + A−η ) · eλt = A−x2.
2. Example 6.3.2
dx
dt= 4x− y, (6.3.3)
dy
dt= x+ 2y,
pA(λ) = λ2 − 6λ+ 9 = 0, λ = 3, 3.
Then we want to find −v2 and −v1 .
A− 3I =
(
1 −11 −1
)
, (A− 3I)2 = 0
We can choose −v2 = (1, 0). Then −v1 = (A− 3I)−v2 = (1, 1). The general solution is
−x = c1
(
1
1
)
e3t + c2
((
1
1
)
t+
(
1
0
))
e3t.
Recommended