High Order Differential

Equations

National Chiao Tung University

Chun-Jen Tsai

10/5/2011

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Second-Order Linear Equations

� A 2nd-order DE of y(x) can be expressed as

G(x, y(x), y'(x), y"(x)) = 0, ∀x ∈ I,

where I is the (open) interval of definition.

� The DE is linear if the equation can be re-written as:

A(x)y" + B(x)y' + C(x)y = F(x), ∀x ∈ I,

In general, we assume that A(x), B(x), C(x), and F(x)

are continuous on the open interval I.

� Recall that if F(x) = 0, the DE is a homogeneous DE.

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Modeling with Linear 2nd-Order DE

� Linear 2nd-order DE are often used to model mechanical systems and electrical systems.

Spring force:FS = –kx, k > 0.

Dashpot force:FR = –c(dx/dt), c > 0.

Since F = mx" = FS + FR, we have

m

spring

x(t)

x = 0 x > 0

mass

dashpot (damper)

.02

2

=++ kxdt

dxc

dt

xdm

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Superposition Principle

� Theorem: Let y1 and y2 be two solutions of the 2nd-order homogeneous linear DE y" + p(x)y' + q(x)y = 0

on the interval I. If c1 and c2 are arbitrary constants,

then the linear combination

y = c1y1(x) + c2y2(x)

is also a solution on I.

Proof:

The proof is as trivial as simply substituting y into the

DE.

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Existence of a Unique Solution

� Theorem: Assume that p(x), q(x), and f(x) are continuous on the open interval I containing the point

a. Then, given any two numbers b0 and b1, the

equationy" + p(x)y' + q(x)y = f(x)

has a unique solution on I that satisfies the initial

conditionsy(a) = b0, y'(a) = b1.

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Example: y" – 2y' + y = 0, y(0) = 3, y'(0) = 1

� It is easy to verify that y1(x) = ex and y2(x) = xex are

solutions of the DE. Therefore, the general solution

can be expressed as

y(x) = c1ex + c2xex.

andy'(x) = (c1 + c2)e

x + c2xex.

The initial conditions tell usc1 = 3 and c1 + c2 = 1. Hence,

the particular solution is

y(x) = 3ex – 2xex.C2 = –2

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Linear Independence of Functions

� Two functions f and g defined on an open interval Iare said to be linearly independent on I if neither one

is a constant multiple of the other. That is, f ≠ kg for any constant number k.

� Examples:

� sin x and cos x

� ex and e–2x

� x+1 and x2

� …

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General Solutions (1/2)

� Given a DE y" + p(x)y' + q(x)y = 0, can we always find at least two linearly independent solutions y1 and y2?

From existence and uniqueness of the IVP solution, we only need to find y1 and y2 such that

y1(a) = 1, y1'(a) = 0 and y2(a) = 0, y2'(a) = 1

because y1(a) ≠ ky2(a) for any constant k.

� Given an IVP, can the solution always be written asy = c1y1(x) + c2y2(x) for some constants c1 and c2?

If so, we have the general solution of the DE.

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General Solutions (2/2)

� Let the initial condition be y(a) = b0, and y'(a) = b1

Since c1 and c2 are solutions to the linear systems

y(a) = c1y1(a) + c2y2(a) = b0

y'(a) = c1y1'(a) + c2y2'(a) = b1.

The solution exists and is unique if the matrix

is non-singular.

.)()(

)()(

21

21

′′ ayay

ayay

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Wronskian of Two Functions

� Definition: Given two functions f(x) and g(x), the Wronskian of f(x) and g(x) is the determinant

� For example, f = ex and g = xex, the Wronskian is

( ) .)()(

)()()(),(

xgxf

xgxfxgxfW

′′=

( ) ., 2x

xxx

xx

xxe

xeee

xeexeeW =

+=

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Checking Independency of Solutions

� Theorem: Suppose that y1 and y2 are two solutions of

the homogeneous 2nd-order linear DE

y" + p(x)y' + q(x)y = 0

on an open interval I where p and q are continuous.

a) If y1 and y2 are linearly dependent, W(y1, y2) ≡ 0 on I.b) If y1 and y2 are linearly independent, W(y1, y2) ≠ 0

at each point of I.

� In the second case, the solution y = c1y1(x) + c2y2(x)

would be a general solution.

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Generalization to nth-Order

� An nth-order linear DE has the form:

y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x).

The corresponding homogeneous linear DE is

y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = 0.

All the properties discussed for the 2nd-order case can be applied to the general n-th order case.

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nth-Order Superposition Principle

� Theorem: Let y1, y2,…,yn be n solutions of the homogeneous nth-order DE on an interval I. If c1,

c2, …, cn are arbitrary constants, then the linear

combination

y = c1y1(x) + c2y2(x) +… + cnyn(x),

is also a solution on I.

� Example: e–3x, cos 2x, sin 2x are solutions of the 3rd-

order homogeneous DE y(3) + 3y" + 4y' + 23y = 0. Is c1cos 2x + c2 sin 2x also a solution?

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Existence of a Unique Solution

� Theorem: Assume that p1(x), p2(x), …, pn(x), and f(x)

are continuous on the open interval I containing the

point a. Then, given any n numbers b0, b1, …, bn–1, the

nth-order equation

y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x)

has a unique solution on I, given the initial conditions

y(a) = b0, y'(a) = b1 , …, y(n–1)(a) = bn–1.

� Note: If f(x) = 0 and all the initial conditions are zero,

then y(x) = 0 is the unique particular solution.

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Example: y(3) + 3y" + 4y' + 12y = 0

� y(x) = c1e–3x + c2cos 2x + c3sin 2x is a solution to the DE.

If y(0) = 0, y'(0) = 5, y"(0) = –39,then c1=–3, c2 = 3, c3 = –2.

If only partial initial conditions are given, we have

different family of solutions

y(0) undetermined y'(0) undetermined y"(0) undetermined

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Linear Dependence of Functions

� Definition: A set of functions f1(x), f2(x), …, fn(x) is said to be linearly dependent on an interval I if there

exist constants c1, c2,…, cn, not all zero, such that

c1 f1(x)+ c2 f2(x)+…+cn fn(x) = 0.

Otherwise, it’s said to be linearly independent.

� Example: The function f1(x) = sin 2x, f2(x) = sin x cos x, and f3(x) = ex are linearly dependent on R since

1· f1 + (–2)· f2 + 0· f3 = 0.

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Wronskian of n Functions

� We are more interested in linearly independent solutions of DE. How can we verify this?

→ Use the Wronskian function.

� Suppose each of the functions f1(x), f2(x),…, fn(x)

possesses at least n–1 derivatives. The determinant

is called the Wronskian of the functions.

( )

)1()1(

2

)1(

1

21

21

21 ,,,

−−−

′′′=

n

n

nn

n

n

n

fff

fff

fff

fffW

L

MMM

L

L

L

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Example: x3y(3) – x2y" + 2xy' – 2y = 0

� Given three solutions y1(x) = x, y2(x) = x ln x, andy3(x) = x2, and the initial conditions y(1) = 3, y'(1) = 2,

y"(1) = 1, verify that y1, y2, and y3 are linearly

independent and find the particular solution.

Note that ln x is defined for x > 0, then

The particular solution is y(x) = x – 3x ln x + 2x2.

.0

2/10

2ln11

ln 2

≠=+= x

x

xx

xxxx

W

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Independency of Solutions (1/3)

� Theorem: Let y1, y2,…,yn be n solutions of the homogeneous nth-order linear DE

y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = 0

on an interval I, where each pi is continuous. Let

W = W(y1, y2,…,yn).

a) If y1, y2 , …, yn are linearly dependent, W ≡ 0 on I.b) If y1, y2 , …, yn are linearly independent, W ≠ 0

at each point of I.

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Independency of Solutions (2/3)

Proof a):

If y1, y2 , …, yn are linearly dependent, ∃c1, c2,…, cn that are not all zeros such that c1y1(x)+…+cn yn(x) = 0.

Differentiate both sides n times, we have the

equivalent system of linear equations

Since c1 = c2 = … = cn = 0 is also a solution, W ≡ 0 on I.

.

0

0

0

)()()(

)()()(

)()()(

2

1

)1()1(

2

)1(

1

21

21

=

′′′

−−−

MM

L

MMM

L

L

n

n

n

nn

n

n

c

c

c

xyxyxy

xyxyxy

xyxyxy

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Independency of Solutions (3/3)

Proof (b):

If y1, y2 , …, yn are linearly independent, then the only

set of c1,…, cn such that c1y1(x)+…+cn yn(x) = 0 is the

set of all zeros. Then the system

has only the trivial solution ∀x ∈ I → the system is non-singular. Therefore W ≠ 0.

.

0

0

0

)()()(

)()()(

)()()(

2

1

)1()1(

2

)1(

1

21

21

=

′′′

−−−

MM

L

MMM

L

L

n

n

n

nn

n

n

c

c

c

xyxyxy

xyxyxy

xyxyxy

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Fundamental Set of Solutions

� Definition: Any set y1, y2, …, yn of n linearly

independent solutions of the homogeneous linear nth-order DE on an interval I is said to be a

fundamental set of solutions on the interval I.

� Let y1, y2,…,yn be a fundamental set of solutions of the homogeneous linear nth-order DE on an interval I.

Then, the general solution of the equation on I is

y = c1y1(x)＋c2y2(x)＋…＋cnyn(x),

where c1, c2,…, cn are arbitrary constants.

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Non-homogeneous Solutions (1/2)

� Theorem: Let yp be any particular solution of the non-homogeneous linear nth-order DE

y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x)

on an interval I, and let y1,y2,…, yn be a fundamental

set of solutions. Then the general solution of the equation on I is:

y = c1y1(x)＋c2y2(x)＋…＋cnyn(x)＋yp,

where c1, c2,…, cn are arbitrary constants.

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Non-homogeneous Solutions (2/2)

Proof:

Let yp(x) be a particular solution of the DE and Y(x) be

any possible particular solution of the DE.

Let yc(x) = Y(x)－yp(x), we have

yc(n) + p1yc

(n–1) + … + pn–1yc' + pnyc

= [Y(n) + p1Y(n–1) + … + pn–1Y' + pnY]

– [yp(n) + p1yp

(n–1) + … + pn–1yp' + pnyp]

= f(x) – f(x) = 0.

→ yc(x) is a solution to the relating homogeneous DE.→ Y(x) = yc(x) + yp(x).

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Complementary Function

� Definition: The general solution of a homogeneous linear nth-order DE is called the complementary

function for the associated non-homogeneous DE.

Let yc(x) = c1y1(x)＋c2y2(x)＋…＋cnyn(x), the general

solution of a non-homogeneous linear nth-order DE

has the form:y(x) = yc(x)＋yp(x).

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Example: y" + 4y = 12x

� It is easy to verify that for the DE, yp = 3x is a particular solution and yc(x) = c1cos 2x + c2sin 2x is its

complementary solution.

If the initial condition is y(0) = 5, y'(0) = 7, since

y(x) = c1cos 2x + c2sin 2x + 3x,

y'(x) = –2c1sin 2x + 2c2cos 2x + 3.

Hence, c1 = 5 and c2 = 2.

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Reduction of Order†

� For a 2nd order linear DE, one can construct a 2nd

solution y2 from a known nontrivial solution y1. If y1

and y2 are linearly independent, we must have

y2/y1 ≠ constant,

Therefore, y2(x) = u(x)y1(x). Substitute this y2(x) into

the DE and solve for u(x) is called reduction of order.

†See problem 36 in section 2.2, page 124 in the textbook

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Example: y" – y = 0

� Solution:Given y1(x) = ex, let y2(x) = u(x) ex,

→ y' = uex + exu', y" = uex + 2exu' + exu"

→ y" – y = ex(u"+ 2u') = 0

→ u"+ 2u' = 0

Let w = u', the DE becomes w' + 2w = 0. Multiplying

by the integrating factor e2x, we have d[e2xw]/dx = 0.Therefore, w = c1e

–2x or u' = c1e–2x.

→ u = (–1/2) c1e–2x + c2.

→ y2(x) = u(x) ex = (–c1/2) e–x + c2ex, let c1 = –2, c2 = 0.

→ Check W(ex, e–x)≠0

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General Case

� Put the 2nd order DE into standard form:

y" + P(x)y' + Q(x)y = 0,

where P(x) and Q(x) are continuous on some interval

I. If y1 is a known solution on I and that y1(x)≠0 for

all x ∈ I, we have:

.)(

)(2

1

)(

12 dxxy

exyy

dxxP

∫∫

=

−

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Example: x2y" – 3xy' + 4y = 0

� Solution:y1 = x2 is a known solution.

→

The general solution is y = c1x2＋c2x

2lnx.

xx

x

dxx

xeedxx

exy

yx

yx

y

xxdxxdx

ln

043

2

2

3ln/3

4

/3

2

2

2

3

=

=

==∫←∫

=

=+′−′′

∫

∫

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Linear Constant Coefficients DEs

� A homogeneous linear nth-order constant coefficient

DE has the form:

any(n) + an–1y

(n–1) + … + a2y" + a1y' + a0y = 0,

where a0, a1, …, an are constant coefficients and an ≠ 0.

� Recall that for a1y' + a0y = 0, the non-singular family of

solution is y = cerx on (–∞, ∞), where r = –a0/a1.

→ Is it possible that for higher order linear constant coefficient DE also has solution in exponential form?

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Characteristic Equations

� Note that if you assume y = erx as a possible solution,

and substitute it into the DE, we have

anrnerx + an–1r

n–1erx + … + a2r2erx + a1rerx + a0e

rx = 0.

Therefore, y = erx is a solution if

anrn + an–1r

n–1 + … + a2r2 + a1r + a0 = 0.

The equation is called the characteristic equation or auxiliary equation of the homogeneous constant

coefficient linear DE.

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Example: 2nd-Order Cases

� Consider a 2nd-order DE, ay" + by' + cy = 0.

Let y = erx, and substituting y' = rerx and y" = r2erx into

the DE, we have: ar2erx + brerx + cerx = 0.

erx > 0 for x ∈ R → ar2 + br + c = 0.

The roots of the characteristic equation gives us the general solution of the DE.

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Classify 2nd-Order Roots (1/2)

� Case I, b2 – 4ac > 0:

r has two real roots r1 and r2, and y1 = er1x and

y2 = er2x form a fundamental set of solutions.

The general solutions is

� Case II, b2 – 4ac = 0:

r has one real root r1 and y1 = er1x. By variation of

parameters (or reduction of order), the 2nd solution of the DE is y2 = xer1x. The general solution is

.21

21

xrxrececy +=

.11

21

xrxrxececy +=

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Classify 2nd-Order Roots (2/2)

� Case III, b2 – 4ac < 0:

r has two complex roots r1 = α + iβ and r2 = α – iβ. Similar to Case I, the general solution is:

� By proper selection of c1 and c2, and using Euler’s

formula, it can be shown that a general solution can

also be represented by

.)(

2

)(

1

xixiececy

βαβα −+ +=

).sincos( 21 xcxceyx ββα +=

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Example: 4y"+4y'+17y = 0

� Solve the IVP: y(0) = –1, y'(0) = 2.

Solution:The roots of the auxiliary equation 4m2+4m+17 = 0 are

m1 = –½ + 2i and m2 = –½ – 2i

→ y = e–x/2 (c1cos 2x + c2sin 2x), andy' = (–c1/2 + 2c2)e

–x/2cos 2x + (–c2/2 – 2c1)e–x/2sin 2x

→ y = e–x/2 (– cos 2x + ¾ sin 2x)

y → 0, as x → ∞.x1

y

1

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Higher-Order Auxiliary Equations

� In general, to solve

any(n) + an–1y

(n–1) + … + a2y" + a1y' + a0y = 0,

where ai ∈ R and an ≠ 0, we must solve

anrn + an–1r

n–1 + … + a2r2 + a1r + a0 = 0.

The general solution of the DE is:

Case I (no repeated roots):

Case II (with repeated roots):

....21

21

xr

n

xrxr necececy +++=

....... 2111

1

1

21

xr

n

xr

k

xrk

k

xrxr knececexcexcecy −++++++= +−

solution form ofrepeated roots

solution form ofdistinct roots

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Differential Operators

� The symbol D, defined by Dy = dy/dx, is called a differential operator. D transforms a function into

another function. For example:

D(cos 4x) = –4sin 4x, D(5x3 – 6x2) = 15x2 – 12x

In general, we define Dny = dny/dxn.

� Polynomial expressions involving D, such as D + 3

and D2 + 3D – 4 are also differential operators. In

particular, they are called polynomial differential

operators.

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nth-Order Differential Operator

� Definition: An nth-order differential operator is:

L = anDn＋an–1D

n–1＋…＋a1D＋a0.

Note that a DE can be expressed in L. In particular,

Ly = 0 stands for

Ly = any(n) + an–1y

(n–1) + … + a1y' + a0y = 0.

� Example: y" + 5y' + 6y = 5x – 3 can be written asD2y＋5Dy＋6y = 5x－3, or (D2＋5D＋6)y = 5x－3.

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Properties of L

� L is a linear operator. That is,

L{α f(x)＋β g(x)}＝α L(f(x))＋β L(g(x)).

� L is also commutative. For example,(D – a)(D – b)y = (D – b)(D – a)y.

Proof:

(D – a)(D – b)y = (D – a)(y' – by)

= D(y' – by) – a(y' – by)

= y" – by' – ay' + aby

= (D – b)Dy – (D – b)ay

= (D – b)(Dy – ay) = (D – b)(D – a)y.

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Solution of Repeated Roots (1/2)

� For a 2nd-order DE, the general solution when the auxiliary equation has repeated roots can be

obtained using variation of parameters.

For the nth-order case, assuming that the auxiliary

equation ofany

(n) + an–1y(n–1) + … + a1y' + a0y = 0

has k repeated roots r0. This means that the DE can

be expressed as (D – r0)k(D – r1) … (D – rn–k)y = 0.

Hence, the solution of (D – r0)ky = 0 will also be a

solution of the nth-order DE.

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Solution of Repeated Roots (2/2)

� Since y1 = er0x is a solution of (D – r0)ky = 0, let

y(x) = u(x)er0x.

Note that(D – r0)[u(x)er0x] = (Du(x))er0x.

Applying the operator k times on y(x), we have

(D – r0)k[u(x)er0x] = (Dku(x))er0x for any u(x).

u(x)er0x is a solution of the DE ↔ Dku(x) = 0.

Possible u(x) that meets this condition is a polynomial

with degree less than k.

→ y(x) = (c1+c2x+ … +ckxk–1)er0x is a family of solutions.

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Example: (D2 + 6D + 13)2y = 0

� The characteristic equation is

(r2 + 6r + 13)2 = 0.

By completing the square, we have

[(r + 3)2 + 4]2 = 0.

The roots are –3±2i of multiplicity 2. Hence the general solution is

y(x) = e–3x(c1cos 2x+d1sin 2x) + xe–3x(c2cos 2x+d2sin 2x).

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Euler Equation†

� Any linear differential equation of the form

where the coefficients ai are constants, is called an

Euler equation.

� Note that anxn = 0 at x = 0. Therefore, we focus on

solving the equation on (0, ∞).

,0011

11

1 =++++−

−−

− yadx

dyxa

dx

ydxa

dx

ydxa

n

nn

nn

nn

n L

†See problem 51 in section 2.1, page 113 in the textbook

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Example: ax2y″ + bxy′ + cy = 0

� If x > 0, the substitution v = ln x transforms Euler

equation into a constant-coefficient linear equation:

with independent variable v. If r1 and r2 are distinct

roots of the characteristic equation, we have a general solution y(x) = c1x

r1 + c2xr2.

� Alternatively, we can substitute y = xm into an Euler

equation and find an alternative way to solve it.

,0)(2

2

=+−+ cydv

dyab

dv

yda

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Alternative Solution to Euler Equation

� Assume that y = xm is a solution, we have

.)1)...(2)(1(

)1)...(2)(1(

...

)1()1( 2

22

22

22

2

1

1

11

00

m

k

kmk

k

k

kk

k

mm

mm

m

xkmmmma

xkmmmmxa

dx

ydxa

xmmaxmmxadx

ydxa

mxamxxadx

dyxa

xaya

+−−−=

+−−−⋅=

−=−⋅=

=⋅=

=

−

−

−

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Example:

� In this example, n = 3, we have the auxiliary equation

am(m–1)(m–2) + bm(m–1) + cm + d = 0,

where a = 1, b = 5, c = 7, and d = 8.

The auxiliary equation then becomes:

m3 + 2m2 + 4m + 8 = 0.

Therefore, m = –2, ±2i are distinct roots. Again, complex exponentials can be removed by Euler’s formula.

08752

22

3

33 =+++ y

dx

dyx

dx

ydx

dx

ydx

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Mechanical Vibrations

� Many dynamic systems can be approximated using a 2nd-order linear DE with constant coefficients:

mx" + cx' + kx = F(t),

where F(t) is the external input force to the system,

and cx' is the damping force of the system.

The behavior of the system can be predicted by the solution x(t) to an IVP x(t0) = x0, x'(t0) = x1 on an

interval containing t0.

m

spring

x(t)

x = 0 x > 0

mass dashpot (damper)

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Free Undamped Motion

� Hooke’s law gives the restoring force of a spring as

F = ks.

Newton’s 2nd law describes the motion of a mass as

m(d2x/dt2) = –k(s + x) + mg= –kx + (mg – ks) = –kx.

System of free undamped motion is

where ω0 = (k/m)½, the solution is

x(t) = A cos ω0t + B sin ω0t.

,)0(,)0(,0 10

2

02

2

xxxxxdt

xd=′==+ ω

equilibrium

position

s

l

m

mmg - ks = 0

motion

x

l + s

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Alternative Form of Solution

� By applying trigonometric formula, we have:

� The motion is called a simple harmonic motion.

A

BBACtCtx =+=−= ααω tan,),cos()( 22

0

x negative

x = 0

x positive

t

ω0: natural frequency of the system.

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Aging Spring

� In real world, the spring constant k usually varies as the spring gets old. Replace k with k(t) = ke–αt, k > 0,

α > 0, we have a more realistic system model:

mx" + ke–αtx = 0.

→ Do you know how to solve this equation?

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Free Damped Motion

� DE of free damped motion:

→

→ The roots of the auxiliary eq.:

dt

dxkx

dt

xdm β−−=

2

2

02 2

02

2

=++ xdt

dx

dt

xdωλ

2

0

2 ωλλ −±−=mm

m

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Three Cases of Damped Motion

� Case I: Over-damped

� Case II: Critically damped

� Case III: Under-damped

)()(20

220

2

21

tttececetx

ωλωλλ −−−− +=

)()( 21 tccetxt += −λ

)sin

cos()(

22

02

22

01

tc

tcetxt

λω

λωλ

−+

−= −

t

x

x

t

x

t

undampedunderdamped

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Non-homogeneous Linear DE

� To solve a non-homogeneous linear DE

any(n) + an-1y

(n–1) + … + a2y" + a1y' + a0y = f(x),

we must do two things:

(1) Find the complementary function yc;(2) Find any particular solution yp of the DE.

→ two methods:� Method of undetermined coefficients

� Variation of parameters

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Undetermined Coefficients

� The method of undetermined coefficients can be applied under two conditions:

1. ai, i = 0, 1, …, n, are constants, and2. f(x) is a linear combination of functions of the

following types:

P(x) = pnxn + pn–1x

n–1 + … + p2x2 + p1x + p0,

P(x)eαx,

P(x)eαxsin βx,

P(x)eαxcos βx.

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Example: y" + 4y' – 2y = 2x2 – 3x + 6

� By guessing, let yp = Ax2 + Bx + C, we haveyp' = 2Ax + B, and yp" = 2A.

Therefore:

yp" + 4yp' – 2yp

= 2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C

= – 2Ax2 + (8A – 2B)x + (2A + 4B – 2C)

= 2x2 – 3x + 6.

→ yp = – x2 – (5/2)x – 9.

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Example: y" – y' + y = 2 sin 3x

� By guessing, let yp = A cos 3x + B sin 3x, we haveyp' = – 3A sin 3x + 3B cos 3x, and

yp" = – 9A cos 3x – 9B sin 3x.

Therefore:

yp" – yp' + yp

= (– 9A – 3B + A) cos 3x + (– 9B + 3A + B) sin 3x

= 2 sin 3x.

→ yp = (6/73) cos 3x – (16/73) sin 3x.

58/82

Example: yp by Superposition

� Solve y" – 2y' – 3y = 4x – 5 + 6xe2x.

By superposition principle, we divide the problem into

two sub-problems, that is,

f(x) = f1(x) + f2(x),

where f1(x) = 4x – 5, and f2(x) = 6xe2x.

By guessing, let yp1 = Ax + B, and yp2 = Cxe2x + Ee2x.Substitute yp = Ax + B + Cxe2x + Ee2x into the DE, we

have:yp = –(4/3)x + (23/9) – 2xe2x – (4/3)Ee2x

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Example: A Glitch in the Method

� Solve y" – 5y' + 4y = 8ex.Simply guessing that yp = Aex and substituting yp into

the DE gives us 0 = 8ex. What went wrong?

Note, if the assumed form of yp is in the solution

space of yc (in this case yc = c1ex + c2e

4x), then we always get 0 = f(x).

Solution, let yp = Axex. Since the derivatives of yp

contains both the term Aex and Axex, it is a reasonable

guess for a particular solution.

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Summary of Two Cases (1/2)

� Case I:No functions in the assumed particular solution is a

solution of the associated homogeneous DE.

→ Substitute with yp = “the form of f(x)”.

A

Ax3＋Bx2＋Cx＋E

A cos 4x＋B sin 4x

Ae5x

(Ax2＋Bx＋C)e5x

Ae3xcos4x＋Be3xsin4x

(Ax2＋Bx＋C)cos4x＋(Ex2＋Fx＋G)sin4x

(Ax＋B)e3xcos4x＋(Cx＋E)e3xsin4x

1. 1 (any constant)

2. x3－x＋1

3. sin4x, or cos4x

4. e5x

5. x2e5x

6. e3xsin4x

7. 5x2sin4x

8. xe3xcos4x

yp

f(x)

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Summary of Two Cases (2/2)

� Case II:A function in the assumed particular solution is also a

solution of the associated homogeneous DE.

→ Substitute with yp = xn × “the form of solution for f(x)”, where n is the smallest positive integer so that yp

is not in the solution space of yc.

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Examples:

� Case I

� y" – 8y′ + 25y = 5x3e–x – 7e–x

� y" + 4y = x cos x

� y" – 9y′ + 14y = 3x2 – 5 sin 2x + 7xe6x

� Case II

� y" – 2y′ + y = ex

� y" + y = 4x + 10 sin x, y(π) = 0, y′(π) = 2

� y" – 6y′ + 9y = 6x2 + 2 – 12 e3x

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Variation of Parameters (1/3)

� Given a 2nd-order DE in standard form:

y" + P(x)y' + Q(x)y = f(x).

We can seek a particular solution using variation of

parameters by assuming that the solution has the

formyp = u1(x)y1(x) + u2(x)y2(x),

where y1 and y2 form a fundamental set of solutions

on I of the associated homogeneous DE.

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Variation of Parameters (2/3)

� Take the derivatives yp' and yp", and substitute them

into the DE, we have

[ ] [ ][ ]

[ ] [ ] [ ]

[ ] [ ] ).()(

)(

)(

)()()()(

)()(

221122112211

221122112211

2211221122221111

22221111

xfuyuyuyuyxPuyuydx

d

uyuyuyuyxPuydx

duy

dx

d

uyuyuyuyxPyuuyyuuy

yxQyxPyuyxQyxPyu

yxQyxPy ppp

=′′+′′+′+′+′+′=

′′+′′+′+′+′+′=

′′+′′+′+′+′+′′+′+′′+

+′+′′++′+′′=

+′+′′

We want this to be zero!

65/82

Variation of Parameters (3/3)

� The solution of the system

can be expressed in terms of determinants:

and

where

W

xfy

W

Wu

)(211 −==′

=′′+′′

=′+′

)(

0

2211

2211

xfuyuy

uyuy

,)(12

2W

xfy

W

Wu ==′

.)(

0,

)(

0,

1

1

2

2

2

1

21

21

xfy

yW

yxf

yW

yy

yyW

′=

′=

′′=

66/82

Summary of the Method

� To solve y" + P(x)y' + Q(x)y = f(x):

� Find yc = c1y1(x) + c2y2(x).

� Find u1 and u2 by integrating u1' = W1/W and u2' = W2/W.

� A particular solution is yp = u1(x)y1(x) + u2(x)y2(x).

� The general solution is y = yc + yp.

� Note that there is no need to introduce any constants when integrating u1' and u2' (why?).

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Examples: y" + y = tan x

� Note that the complementary function is

yc(x) = c1 cos x + c2 sin x.

Compute W, W1, W2 and obtain

W = cos2x + sin2x = 1,

W1 = – sin 2x / cos x = cos x – sec x,W2 = sin x,

→ u1' = cos x – sec x, and u2' = sin x.

→ u1 = sin x – ln |sec x + tan x|, and u2 = –cos x.→ yp(x) = (sin x – ln |sec x + tan x|) cos x – cos x sin x.

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High-Order Variation of Parameters

� For a linear nth-order DE

y(n) + Pn–1(x)y(n–1) + …+ P1(x)y' + P0(x)y = f(x),

if yc = c1y1 + c2y2 + … + cnyn is the complementary

function of the DE, then a particular solution is

yp = u1(x)y1(x) + u2(x)y2(x) + … + un(x)yn(x),

where uk′ = Wk/W, k = 1, 2, …, n and W is the Wronskian of y1, y2, .., yn and Wk is the determinant

obtained by replacing the kth column of the Wronskian with the column (0, 0, …, f(x))T.

69/82

Forced Oscillation and Resonance

� Now, consider the effect of externalforce f(t) on the damped motion system:

→

Note: ω0 is the natural frequency of the system.

)(2

2

tfdt

dxkx

dt

xdm +−−= β

)(2 2

02

2

tFxdt

dx

dt

xd=++ ωλ

m

moves theceiling ofthe platform

70/82

Transient and Steady-State Terms

� When F(t) is a periodic function and λ > 0, the solution is the sum of a non-periodic function xc(t) and

a periodic function xp(t). Moreover limt→∞ xc(t) = 0.

1

-1

transient

steady-state

x

t

1

-1

x( t) = transient

+ steady-state

x

t

π/2 π/2

xp(t)

71/82

Example: Transient/Steady State

� The solution of the IVP

isx(t) = (x1 – 2)e–t sin t + 2 sin t,

12

2

)0(,0)0(,sin2cos422 xxxttxdt

dx

dt

xd=′=+=++

transient steady-state

4

- 4

x1

= 7

x

t

- 2

2

x1

= 3

x1

= 0

x1

= -3

2ππ

72/82

Undamped Forced Motion

� The solution of

is

where c1 = 0, c2 = –γF0 /ω0(ω02 –γ2).

→

→ There is no transient term.

0)0(,0)0(,sin0

2

02

2

=′==+ xxtFxdt

xdγω

,sinsincos)(22

0

00201 t

Ftctctx γ

γωωω

−++=

.,)sinsin()(

)( 00022

00

0 ωγγωωγγωω

≠+−−

= ttF

tx

73/82

Pure Resonance

� In the previous example, when γ → ω0, the displacement of the system become large as t → ∞.

)(

)sinsin(

lim

)(

sinsinlim)(

2

0

3

0

00

0

22

00

000

0

0

γωωγ

γωωγγ

γωω

γωωγ

ωγ

ωγ

−

+−

=

−

+−=

→

→

d

d

ttd

d

F

ttFtx

.cos2

sin2

0

0

002

0

0 ttF

tF

ωω

ωω

−=

t

x

74/82

Tacoma Narrow Bridge, WA, USA

� Opened in July 1, 1940, collapsed in Nov. 7, 1940.

� The wind-blow frequency matched the natural frequency of the bridge, which caused a pure resonance effect that destroyed the bridge.

75/82

Damping System of Taipei 101

� Taipei 101 uses a 730-ton damping ball to stabilize the building

76/82

Boundary-Value Problem

� Solving a linear DE with y or its derivatives specified

at different points. For example, solve

y" + P(x)y' + Q(x)y = f(x),

subject toy(a) = y0, y(b) = y1.

Such problems are called end-pointor boundary-value problems (BVP).

y

x

I

(a, y0)

(b, y1)

solutions of the DE

77/82

Example: y" + 16y = 0

� One can verify that y = c1 cos 4x + c2 sin 4x is a family of solutions of y" + 16y = 0. What are the solutions of

the BVPs with

� y(0) = 0, y(π/2) = 0?

� y(0) = 0, y(π/8) = 0?

� y(0) = 0, y(π/2) = 1?

y

x

( /2, 0)(0, 0)

1

–1

π

02 =c

21

2 −=c

12 =c

41

2 =c

21

2 =c

78/82

Eigenvalue Problems

� An eigenvalue problem in DE is a homogeneous BVP such that the boundary conditions evaluate to 0 and

there is a parameter λ at the coefficient of y. That is

y" + p(x)y' + λq(x)y = 0, y(a) = 0, y(b) = 0.

The eigenvalue problem tries to find a λ (eigenvalue) such that the BVP has a nontrivial solution. The non-

trivial solution is then called an eigenfunction.

� In last example, p(x) = 0, λ = 16, q(x) = 1, a = 0, b = π/2. The 2nd BVP has only trivial solution.

→ what is the eigenvalue λ?

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Example: y" + λy = 0, y(0) = y(L) = 0

� The problem can be solved by enumerating different

cases when λ = 0, λ < 0, and λ > 0.

(1) λ = 0, we have y″ = 0,→ the general solution is y(x) = Ax + B.

→ y = 0 is the only solution for the BVP→ λ = 0 is not an eigenvalue of the BVP

(2) λ < 0, let λ = –α2, α > 0, we have y″ – α2y = 0,

→ the general solution is y(x) = c1eαx + c2e

–αx.

→ y = 0 is the only solution for the BVP→ λ = 0 is not an eigenvalue of the BVP

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Example: y" + λy = 0, y(0) = y(L) = 0

(3) λ > 0, let λ = α2, α > 0, we have y″ + α2y = 0,→ the solution is y(x) = c1cos(αx) + c2 sin(αx).

→ y(0) = 0 implies c1 = 0

→ y(L) = 0 implies sin(αL) = 0, or αL = nπ, n∈ Z

→ The BVP has infinitely many eigenvalues:

...,3,2,1,2

22

== nL

nn

πλ

81/82

The Deflection of a Uniform Beam

� The deflection of a curve can be represented by a 4th-order DE:

)(4

4

xFdx

ydEI =

load per unit lengthflexural rigidity

A flexible beam The deflection curve

82/82

Boundary Conditions

� Endpoint conditions of a flexible beam:

y″′ = 0y″ = 0Free end

y″ = 0y = 0Simply supported

y′ = 0y = 0Embedded

Endpoint conditionsEnd of beam

x = 0 x = L x = 0 x = L x = 0 x = L

Embedded at both ends Free at the right end Supported at both ends