1

Differential Equations

1. Introduction

An equation involving one or more derivatives of an unknown function is called a differential

equation. Differential equations arise in many physical phenomena and mathematical analysis of

any engineering problems.

A mathematician is interested in proving that a given differential equation possesses a solution;

hence one can obtain the solution and deduce a few properties of that solution. A physicist or an

engineer on the other hand is usually interested in the specific expression of the solution. The

usual compromise is to find the solution.

1.1 Fundamental definitions

An ordinary differential equation is an equation which involves derivatives of an unknown

function y of a single variable x.

Ordinary differential equations (ODEs) arise in many different contexts throughout mathematics

and science (social and natural) one way or another, because when describing changes

mathematically, the most accurate way is differentials and derivatives (related, though not quite

the same). Since various differentials, derivatives, and functions become inevitably related to

each other via equations, a differential equation is the result, describing dynamical phenomena,

evolution, and variation. Often, quantities are defined as the rate of change of other quantities

(time derivatives), or gradients of quantities, which is how they enter differential equations.

Specific mathematical fields include geometry and analytical mechanics. Scientific fields include

much of physics and astronomy (celestial mechanics), geology (weather modeling), chemistry

(reaction rates), biology (infectious diseases, genetic variation), ecology and population

modeling (population competition), economics (stock trends, interest rates and the market

equilibrium price changes). Many mathematicians have studied differential equations and

contributed to the field, including Newton, Leibniz, the Bernoulli family, Riccati, Clairaut,

d'Alembert, and Euler. A simple example is Newton's second law of motion — the relationship

between the displacement x and the time t of the object under the force F, which leads to the

differential equation.

Examples:

The order of a differential equation is the order of the highest derivative occurring in it.

The degree of a differential equation is the degree of the highest derivative exists in the

equation, provided the equation is free of radicals and terms with fractional degree.

The order and degree of the differential equations in the above examples are respectively

2

A partial differential equation is an equation which involves partial derivatives of unknown

functions of two or more independent variables.

Here, we consider only ordinary differential equations.

A differential equation is said to be linear if it is a linear function of the dependent variable and

its derivatives, i.e., it is of degree one in the dependent variable y and its derivatives.

The general linear differential equation of order n is of the form

Where b0, b1,…….. bn, and R(x) are functions of x alone.

A solution of a differential equation is a relation between the dependent and independent

variables, having derivative of order equal to order of the differential equation ,which is also free

of the derivatives and satisfying the given differential equation.

Example 1.1

The solution of is

Since .

This solution is a particular solution. The expression also satisfies the equation and it is

the general solution of this equation, for any arbitrary c.

Example 1.2

Consider the differential equation 2

23 2 0

d y dyy

dxdx .

Let y = e2x

, then 2

2 2

22 4 .x xdy d ye and e

dx dx

Hence 2

23 2 0

d y dyy

dxdx implying that y = e

2x is a solution of the given differential equation.

Also y = ex is a solution. These two solutions are particular solutions of the given equation. Note

that y = C1 e2x

+ C2 ex is also a solution , where C1and C2 are arbitrary constants. This solution is

called the general solution of the given differential equation, which is the linear combination of

all possible linearly independent solutions.

Note:

A differential equation together with an initial condition is called an Initial Value problem. The

initial condition is used to determine the value of the arbitrary constants in the general solution.

1.2 Formulation of differential equations by eliminating arbitrary constants

In practice, differential equations arise in many ways, one of which is useful in that it gives us a

feeling for the kinds of solutions to be expected. In this section we start with the relation

3

involving arbitrary constants, and, by elimination of those arbitrary constants obtain a differential

equation which is consistent with the original relation. In other words we will obtain a

differential equation for which the given relation is the general solution.

Methods for elimination of arbitrary constants vary the way in which the constants enter the

given relation. Since each differentiation yields a new relation, the number of derivatives that

needs to be used is same as that of the number of arbitrary constants to be eliminated. Thus in

eliminating arbitrary constants from a relation we obtain a differential equation that is

(i) Of order equal to the number of arbitrary constants in the equation.

(ii) Consistent with relation.

(iii)Free from arbitrary constants.

Example 1.2.1 Eliminate the arbitrary constants c1 and c2 from the relation

2 3

1 2 . (1)x xy c e c e

Since two constants are to be eliminated, obtain the two derivatives, 2 3

1 2

2 3

1 2

2 3 , (2)

4 9 . (3)

x x

x x

y c e c e

y c e c e

The elimination of 1 c from equations (2) and (3) yields

3

22 15 ;xy y c e

the elimination of 1 c from equations (1) and (2) yields

3

22 5 .xy y c e

Hence 2 3( 2 ),or 6 0.y y y y y y y

Another method for obtaining the differential equation in this example proceeds as follows. We

know from a theorem in elementary algebra that the equations (1), (2), and (3) considered as

equations in the two unknowns 1 2 and c c can have solutions only if

2 3

2 3

'' 2 3

2 3 0. (4)

4 9

x x

x x

x x

y e e

y e e

y e e

Since e-2x

and e3x

cannot be zero, equation (4) may be rewritten, with the factors e-2x

and e3x

removed, as

1 1

2 3 0

4 9

y

y

y

from which the differential equation 6 0y y y follows immediately.

4

This latter method has the advantage of making it easy to see that the elimination of the constants

1 2 , ,..., nc c c from a relation of the form 1 2m m

1 2e + e ... nm xx x

ny c c c e will always lead to a

differential equation 1

0 1 1... 0,

n n

nn n

d y d ya a a y

dx dx

in which the coefficients 0 1, ,..., na a a are constants. The

study of such differential equations will receive much of our attention.

Example 1.2.2 Eliminate the constant a from the equation 2 2 2( ) .x a y a

Direct differentiation of the relation yields 2( ) 2 0,x a yy from which .a x yy

Therefore, using the original equation, we find that 2 2 2 ,y x xyy which may be written in

the form 2 2( ) 2 0.x y dx xydy

Another method is based upon the isolation of an arbitrary constant.

The equation 2 2 2( )x a y a may be put in the form

2 2

2 .x y

ax

Differentiating both sides, we get 2 2

2

(2 2 ) ( )0,

x xdx ydy x y dx

x

or

2 2( ) 2 0,x y dx xydy as desired.

Example 1.2.3 Eliminate and B from the relation cos( ),- - - - - - - - (5)x B t in

which is a parameter (not to be eliminated).

First we obtain two derivatives of x with respect to t

22

2

sin( ), (6)

cos( ). (7)

dxB t

dt

d xB t

dt

On comparing equations (5) and (7) we get 2

2

20.

d xx

dt

Example 1.2.4 Eliminate c from the equation 2 4 0.cxy c x

Differentiating the given equation we get 2( ) 0.c y xy c

Since 0, ( )c c y xy and substitution into the original leads to the result

3 2 2( ) 4 0.x y x yy

Exercises 1.2.5

Form the differential equation by eliminating the arbitrary constants

1. 1 2cos sin ; a parameter.x c t c t

5

2. 2 4 .y ax

3. 2 2

1 2 .x xy x c e c e

4. 2 2 .x xy Ae Bxe

5. 1 2cos sin ; a and b are parameters.ax axy c e bx c e bx

1.3 Families of curves

A relation involving a parameter, as well as one or both the coordinates of a point in a plane,

represents a family of curves, one curve corresponding to each value of the parameter. For

instance, the equation 2 2 2( ) ( ) 2x c y c c may be interpreted as the equation of a family

of circles, each having its center on the line y = x and each passing through the origin.

If the constant c is treated as an arbitrary constant and eliminated, the result is called the

differential equation of the family represented by the equation. In this case, the elimination of c

is easily performed by isolating c, then differentiating throughout the equation with respect to x.

Thus, the given equation takes the form

2 2

2x y

cx y

and hence we can form the differential

equation.

Note that for a two parameter family of curves, the differential equation will be of order 2.

Example 1.3.1 Find the differential equation of the family of parabolas, having their vertices at

the origin and their foci on the y-axis.

From analytic geometry, we find the equation of this family of parabolas to be 2 4 .x ay Then

from F F

M and Nx y

then a may be eliminated by differentiation. Thus it follows

that 22 0.xydx x dy

Example 1.3.2 Find the differential equation of the family of circles having their centers on the

y-axis.

Since a member of the family of circles of this example may have its center anywhere on the y-

axis and its radius of any magnitude, we are dealing with the two-parameter family

2 2 2( ) .x y b r

Differentiating the given equation, we get

( ) 0,x y b y from which .x yy

by

6

Differentiating again,

2

2

[1 ( ) ] ( )0,

( )

y yy y y x yy

y

so the desired differential

equation is 3( ) 0.xy y y

Exercises 1.3.3

In each exercise, obtain the differential equation of the family of plane curves described and

sketch several representative members of the family.

1. Straight lines with slope and x-intercept equal.

2. Circles with fixed radius r and tangent to the x-axis.

3. Parabolas with axis parallel to the x-axis.

4. The cubics 2 2( ) cy x x a

with a fixed.

5. The cissoids

32 .

xy

a x

6. The strophoids (sec tan ).r a

1.4 Differential equations of order one and degree one

In our study we initially consider only first order and first degree differential equation. Such

equations can be written in the form

Before discussing some of the analytic techniques for finding solutions we shall state an

important theorem concerning to the uniqueness and existence of solution.

Consider

Let T denote the rectangular region defined by and , a region with

the point at its centre. Suppose that and are continuous functions of and in

Under the conditions imposed on above, an interval exists about , and

satisfies the properties.

a) A solution of equation (1) in

b) On the interval , satisfies

c) At ,

d) is unique in

satisfying the above conditions.

In other wards, the theorem states that if is sufficiently well behaved near the

point , then the differential equation ,

7

has a solution that passes through the point and that solution is unique near

We consider some first order differential equation in the following forms.

1.5 Variable Separable equations

If the differential equation M(x,y) dx + N(x,y) dy = 0 is simple enough that the variables can be

separated ,i.e., the equation can be rewritten as F(x) dx + G(y) dy = 0, where F is a function of

x alone and G is a function of y alone , then the solution can be obtained by direct integration.

Example 1.5.1

Integrating,

i.e. , is the required solution, where c is some arbitrary constant .

Example 1.5.2 Consider the equation,

Here the variables are not separated but it can be reduced to that form by dividing the equation

by Then

where the variables are separated.

The solution is

Remark: From the above example it is clear that an equation of the type

can be reduced to variable separable equation.

Example 1.5.3

Solution:

The solution is

Example 1.5.4

Solution:

8

Then given D.E becomes, 2 21 1 1 0x y dx x y dy

Example 1.5.5

Solution:

Then

Integrating,

Exercise 1.5.6

Solve

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1.6 Differential equations with homogeneous coefficients:

Consider the differential equation of the form , where and are

homogeneous functions of the same degree in and .

9

To solve this equation, we note that being a homogenous equation of degree zero , is a

function of (y/x) only. Let . This suggests the substitution (y/x) = v or

then . Substituting in the given equation we get = g(v)

or = g(v) – v, which can be solved by separating the variables.

Example 1.6.1

Solution:

Put then

Example 1.6.2

Solution:

then

Integrating,

Remark: It is quite immaterial whether one uses or . However, it is sometimes

easier to solve by substituting for the variable whose differential has the simpler coefficient.

Example 1.6.3

Solution:

10

Integrating,

Exercise 1.6.4

Solve

1.

2.

3.

4.

5.

6.

7.

8.

1.7 Differential Equations with linear coefficients:

The equations of the form can be reduced to the homogenous form.

Case 1: When

Putting x= X + h and y = Y + k, where h, k are constants so that dx = dX , dy = dY

Choose h, k such that ah+bk+c=0 and 0 0a h b k c ab a b

and which is of homogenous coefficients in and can be solved.

Case 2: When i.e., , then equation can be reduced to variable-separable

Form by substituting and hence can be solved..

Example 1.7.1

Solve

Clearly,

11

put

Substituting,

Integrating both sides,

Finally, .

Example 1.7.2 Solve

Here

Put

i.e

i.e

Separating,

i.e Solution is

Exercise 1.7.3

Solve

1.

2.

3.

4.

5.

6.

7.

8.

9.

12

1.8 Exact Equations

These are equations of the type in which separation of variables

may not be possible. Suppose there exists a function such that is ,

then the solution is .

If the equation is exact, then by definition there exists a F such that

But from total derivate formula ,

Comparing the above equations, we get F F

M and Nx y

These two equations lead to and

Since ……………………(1)

Thus for an equation M dx + N dy = 0 to be exact it is necessary that the condition (1) is to be

satisfied. Now let us prove the converse of the above result. Let us assume that the condition (1)

holds. Let (x, y) be a function for which ( , )M x yx

. (i.e., the function is the result of

integrating M dx w.r.t. x alone holding y as a constant.

Then 2

;M

y x y

Hence from (1) ,

2 2 M N

x y y x y x

.

On integrating both sides of this equation w.r.t. x holding y fixed we get,

( ),N B yy

where B(y) is an arbitrary function of y.

Now define a function, ( , ) ( , ) ( ) .F x y x y B y dy

Then ( ) .F F

dF dx dy dx B y dy Mdx Ndyx y x y

Hence the given equation is exact. Thus , we have proved the following theorem.

Theorem: If M , N, M N

andy x

are all continuous functions of x and y , then a necessary and

sufficient condition for the differential equation M dx + N dy = 0 to be exact is that

Note : If the differential equation M dx + N dy = 0 is exact then the solution of the equation can

be obtained as follows. Let F be a function of x and y such that dF = M dx + N dy.

Then comparing with the equation F F

dF dx dyx y

we get F

x

= M (x, y).

13

On integrating w.r.t. x, holding y as a constant we get F(x, y) = ( , ) ( )M x y dx B y , where B(y)

is an arbitrary function of y alone. Now

( , ) ( ) ( , )F

M x y dx B y N x yy y

implies that ( )B y consists of terms in N(x, y) which

does not contain x.

Thus the solution function is given by F(x, y) = ( , ) ( )M x y dx B y

= ( , ) ( ( , ) )M x y dx termsin N x y not containg x dy

Example 1.8.1 Solve

M = 3x2y – 6x ……………..(1)

N = x2 +2y …………………..(2)

and .

Hence the equation is exact. Therefore the solution is given by

( , ) ( ( , ) )M x y dx termsin N x y not containg x dy = c

or,

Exercises 1.8.2:

Solve

1.

2.

3.

4.

5.

1.9 Linear Equations

Definition: A differential equation is said to be linear if the dependent variable and its

differential coefficient occur only in the first degree and not multiplied together.

A general linear differential equation is of the form

To solve, multiply both sides by P dx

e so that

The above equation is equivalent to

Integrating, we get as required solution.

If , then solution can be rewritten as

Example 1.9.1 Solve

Dividing throughout by ,

14

Thus the solution is .

Example 1.9.2 Solve

Which is linear equation in where

Thus the solution is

where

Exercises 1.9.3

Solve

1.

2.

3.

4.

5.

1.10. Bernoulli equation

The equation is reducible to the linear equation and is usually called

Bernoulli’s equation.

To solve, divide both sides by , so that .

Put . Equation reduces to which is linear in and

can be solved.

Example 1.10.1 Solve .

Dividing by , .

Put , so

15

i.e. , i.e which is linear in z.

I.F = = =

Solution is

i.e.

i.e.

Example 1.10.2 Solve

Dividing by ,

1 1

( )1

dxdy

y y x

i.e.

Exercise 1.10.3

Solve

1.

2.

3.

4.

5.

6.

7.

8. ,

1.11 Equations reducible to exact equations

Consider the equation which is not exact..…..(1)

Let , a function of and , be an integrating factor of (1). Then

must be exact.

Hence, must satisfy .

16

.

First let be a function of alone. Then and becomes .

Then we have or

If the left hand side of the above equation as function of alone, we have

Then the desired I.F is

By a similar argument, assuming is a function of alone, we get

Then an integrating factor is .

Using the above integrating factors, one can convert the equation to exact form and solve.

Example 1.11.1 Solve

and

Hence, .

.

Multiply to the above equation, we get

.

Solution is .

Example 1.11.2 Solve .

,

So

I.F =

Multiplying to the equation we get .

We get the solution .

Exercise 1.11.3 Solve

1.

2.

3.

4.

17

5.

6.

7.

8.

1.12 Integrating factors by inspection.

Here we are concerned with the equations that are simple enough to enable us to find the

integrating factors by inspection. The ability to do this depends largely upon recognition of

certain common exact differential and upon experience. Below are four exact differentials that

occur frequently:

Using these exact differentials it is possible to group the terms in given differential equation and

obtain the integrating factors.

Example 1.12.1 Solve .

Group the terms

Dividing by ,

Solution on integration, .

Example 1.12.2 Solve 2cos ( )dy y

x ydx x

2cos ( )y

xdy ydx dxx

2 2( ) cos ( )y y

d x dxx x

2

2sec ( ) ( )

y y dxd

x x x

1tan( )

yc

x x

Example 1.12.3 Solve 2 4 33 ( ) 0x ydx y x dy ,

Rewriting the given differential equation as 2 3 4(3 ) 0x ydx x dy y dy .

Which is nothing but 3 3 4( ) 0.yd x x dy y dy Dividing by y2, the equation becomes,

18

32 0

xd y dy

y

, which is exact. Therefore the solution is

3 3

3

x yC

y .

Exercise 1.12.4

Solve

1.

2.

3.

4.

5.

6.

7.

1.13 Higher Order Linear Differential Equations

The general linear differential equation of order n is an equation that can be written as

……………….(1)

If R(x) = 0, then the equation is called a homogenous linear differential equation otherwise it is

called non-homogeneous differential equation. The coefficient functions are

continuous on the interval I.

Consider .

If and are solutions of homogenous equation then is also a solution of

that equation.

In a similar manner, if y1 ,y2 ,…,yn are solutions of (1), then

……..(2)

is also a solution where are arbitrary constants.

The expression (2) is called linear combination of the functions .

For a homogeneous equation, the

general solution is of the form where are all possible

linearly independent solutions of the given equation. Where as for the nonhomogeneous equation

the general solution is of the

form c py y y , where cy is the general solution of the corresponding homogenous

equation and py is a particular

solution of the given equation , which does not contain any arbitrary constants.

1.3.1 An Existence and Uniqueness Theorem:

Given an order linear differential equation,

19

on an interval I,

suppose that is any number on the interval I and are n arbitrary constants.

Example 1.13.1 Find the unique solution of , .

Solution is .

Using we get .

1.13.2 Linear Independence of Solutions:

Given the functions if constants not all zero, exist such that

for all in , then the functions

are said to be linearly dependent on that interval. If no such relation exists, the functions are said

to be linearly independent.

1.13.3 The Wronskian of Solution:

To test whether n functions are linearly independent on an interval let us assume that

each of the functions is differentiable atleast ( times .

Then from the equation 1 1 2 2 ... n nc f c f c f = 0 , it follows by successive differentiation that

' ' '

1 1 2 2 ... n nc f c f c f = 0

''

1 1 2 2 ... n nc f c f c f = 0

… 1 1 1

1 1 2 2 ...n n n

n nc f c f c f = 0

For any fixed value of in , the nature of solutions of these will be determined by the

determinant

1 2

' ' '

1 2

1 1 1

1 2

( ) ( ) ... ( )

( ) ( ) ... ( )( )

. . . .

( ) ( ) ... ( )

n

n

n n n

n

f x f x f x

f x f x f xW x

f x f x f x

If for some on , then c1 = c2 =…=cn = 0 Hence, functions are linearly

independent on . The function is called Wronskian of the functions .

Example 1.13.2 Let 1 2 1 2, ,ax bx ax bxy e and y e then y ae y be .

( ) 0

ax bxa b x

ax bx

e eThen wronskian is given by W b a e

ae be

only if a=b.

Therefore for a≠b, 1 2ax bxy e and y e are linearly independent.

Example 1.13.3. Let 1 2 1 21 , 0, 1y and y x then y y

20

11

0 1

xThen wronskian is given by W ≠ 0. Therefore 1 21y and y x are linearly

independent.

Let 1,2,....k

k

k

d dD then D for k

dx dx .Then the equation (1) can be expressed as

10 1

10 1

b ... ( )

. ., (b ... ) ( )

n nn

n nn

D y b D y b y R x

i e D b D b y R x

i.e., f(D) y = R(i) where f(D) = 1

0 1b ...n nnD b D b is called a differential operator. To

solve such equations we first study the properties of the differential operator.

1.13.4 Properties of differential operator:

1. f(D)eax

= eax

f(a)

Proof: Let f(D) = 1

0 1b ...n nnD b D b .

Since Dke

ax = a

ke

ax, for k = 1,2,3…n. we have,

ax 10 1

10 1

10 1

f(D)e = b ...

...

( ... ) ( )

n ax n ax axn

n ax n ax axn

n n ax axn

D e b D e b e

b a e b a e b e

b a b a b e f a e

2. f(D)eax

y = eax

f(D+a)y

Proof: Let f(D) = 1

0 1b ...n nnD b D b

We have D eax

y = y Deax

+ eax

Dy = y eax

a + eax

Dy = eax

(D+a)y,

D2(e

ax y) = D(D(e

ax y)) = D(e

ax (D+A)y) = e

ax (D+a)(D+a)y= e

ax (D+a)

2y

Similarly we can show that Dk(e

ax y) = e

ax (D+a)

ky, k = 1,2,...n.

Substituting in the formula we get the required result.

3. 0, 0,1,2,... 1

( )!,

k ax j

ax

j kD a e x

e k j k

Proof: We know that 0, 0,1,2,.... 1

!,

k j j kD x

k j k

then by property(2), the result follows.

1.13.5 The solution of linear homogeneous differential equation with constant coefficients

Consider the linear homogeneous differential equation with constant coefficients

1

0 1 11... 0

n n

n nn n

d y d y dyb b b b y

dx dx dx

where are all constants.

21

This equation can be rewritten as f(D)y = 0 where f(D) = 10 1b ...n n

nD b D b .

If y = axe , then by property (1), f(D)y = f(D)eax

= eax

f(a) = 0 f(a) = 0.

This equation is called the auxiliary or characteristic equation associated with the given

differential equation. For a nth

order differential equation, the auxiliary equation has n roots say,

1 2, ,...... .na a a Then 1 21 2, ,..........., na xa x a x

ny e y e y e are all solutions of the given differential

equation.

Case1: If the roots of the auxiliary equations are all distinct then

1 21 2, ,..........., na xa x a x

ny e y e y e are linearly independent solutions of the given equation.

Hence the general solution is given by 1 21 2 ... na xa x a x

ny C e C e C e .

Case2: If some roots are equal, say 1 2 ...... .ka a a a

Then the solutions 1 2 ... axky y y e . Then the solution is given by 1 2 ... na xa x a x

y e e e

does not contain n arbitrary constants and hence cannot be the general solution. Since first k

roots of the auxiliary equations are equal the given differential equation can be rewritten as

( )( ) 0.kg D D a y Then by property (3), we observe that , 0,1,......,. 1ax jjy e x j k are all

solutions of the given equation. Hence the general solution is

12 11 2 3 1( ...... ) ......k na x a xk ax

k k ny C C x C x C x e C e C e .

Case3. If some roots of the auxiliary equation are real. Since for equation with real coefficients

the roots exists in conjugate pairs, let 1 2a iband a ib be two roots. Then

1 2( ) ( )1 2(cos sin ) (cos sin )

x xa ib x ax a ib x axy e e e bx i bx and y e e e bx i bx are the two

distinct solutions. Therefore the corresponding solution is 1 21 2

x xy C e C e , which can be

expressed as 1 1 2( cos sin )axy e A bx A bx where 1 2A and A are arbitrary constants.

Example 1.13.4: Solve (2D2 + 5D – 12)y = 0

Solution: Auxiliary equation is 2m2 + 5m – 12 = 0.

That is, (2m – 3) (m + 4) = 0 and it has the roots m1 = 3/2 and m2 = - 4 which are real and

distinct. Therefore the general solution is y = C1 e3x/2

+ C2e-4x

Example 1.13.5 Solve 2

2

dx

yd + 4 y = 0

22

Solution: Auxiliary equation is m2 + 4 = 0 or m

2= - 4

Therefore m = 2i = 0 2i

Therefore y = C1 cos 2x + C2 sin 2x

Example 1.13.6 Solve 3

3

dx

xd + 4

2

2

dx

yd + 4

dx

dy = 0.

Solution: Auxiliary equation is m3 + 4m

2 + 4m = 0.

That is, m(m + 2)2 = 0

Therefore m1 = 0, m2 = -2, m3 = -2 are its roots.

Here we find that two roots are equal. Hence the general solution is given by

y = C1 e0x

+ (C2 + C3x) e-2x

Or y = C1 + (C2 + C3i) e-2x

Example 1.13.7 Solve4

4

dx

xd + 8

2

2

dx

yd + 16y = 0

Solution: Here the auxiliary equation is m4 + 8m

2 + 16 = 0, which is simply (m

2 + 4)

2 = 0.

Therefore we have m2 + 4 = 0 repeated. It gives m

2 = -4

Therefore m = 2i.

Thus the roots are m = 2 i, 2i (imaginary, repeated).

Hence the general solution is y = e0x

[(C1 + C2x) cos 2x + (C3 + C4x) sin 2x]

That is, y = (C1 + C2x) cos 2x + (C3 + C4x) sin 2x.

Example 1.13.8 Solve 4

4

dx

yd - 2

3

3

dx

yd + 2

2

2

dx

yd - 2

dx

dy + y = 0

Solution: Hence the A.E. is m4 – 2m

3 + 2m

2 – 2m + 1 = 0.

The roots are m = 1, 1, i

Therefore the general solution is y = (C1 + C2x) ex + e

0x (C3 cos x + C4 sin x)

That is, y = (C1 + C2x)ex + (C3 cos x + C4 sin x).

1.13.6 The solution of linear non-homogeneous differential equation with constant

coefficients

We know that the general solution of nonhomogeneous linear differential equation

23

is of the form c py y y , where cy is the general solution of the corresponding homogenous

equation , called the complementary function and

py is a particular solution of the given equation , which does not contain any arbitrary constants.

The complementary function can be determined using the method described above. To determine

the particular solution py , we use the following methods.

1.13.6.1 Inverse differential operator method

If f(D)y = (x) then we define the inverse differential operator denoted by 1

( )f D as

)(

1

Df [ (x)] = y , where ‘D’ is the differential operator

dx

d.

Thus f(D) is also a differential operator and )(

1

Df can be treated as its inverse.

Example 1.13.6.1 1

( ) ( ) .y ydx DR x y R x ydxD

2

2

1 1 1 1 11 1. , 1 1

2

11 , 1,2,3,.......

!

k

k

xdx x x xdx

D D D DD

xSimilarly wecan showthat k

kD

1.13.6.2. Properties of the inverse differential operator:

2. 1

( ) 0.( ) ( )

axax e

e if f af D f a

Proof: We know that f(D)eax

= eax

f(a), If f(a) 0, then dividing by f(a) we get

1( ) ( )

( ) ( )

1

( ) ( )

axax

axax

ef D e f a

f a f a

ee

f D f a

3. 1 1

( ) ( )

ax axe y e yf D a f D

Proof: From the property f(D)eax

y = eax

f(D+a)y the result follows.

4. 1 ax axy e e y dx

D a

Proof: The result follows directly from the result (2) and example (1) .

24

To determine the particular solution of a linear non-homogenous differential equation, we use

inverse differential operators. If f(D)y = (x) ,then yp = )(

1

Df [ (x)]

Case(i): If (x) = eax

, then yp = ( )

axe

f a if f(a) 0.

If f(a) = 0, then f(D) can expressed as f(D) = (D-a)k (D), where (a) 0.for k = 1,2,3,….

k k k

k k

1 1 1 1 1 Then

f(D) (D) (a)(D-a) (D) (D-a) (D-a)

1 1 11

(a) (a) (a) !(D-a) D

axax ax ax

ax ax kax

ee e e

e e xe

k

Case(ii): If (x)= cosax or sinax then, since cosax = Re(iaxe ) and sinax = Im(

iaxe ) this case

reduces to the case(i) and hence can be solved.

Case(iii) If (x)= xm, for some positive integer m, then

1

f(D) can be expanded as a series in

positive powers of x and hence 1

f(D)

mx can be determined.

Working Rule:

1. If (x) = eax

, then

2

2

1( ) 0.

( ) ( )

1 1( ) 0 ( ) 0,

( ) ( ) ( )

1( ) 0.

( )

1( ) 0 ...

( )

axax

axax ax

ax

ax

ee if f a

f D f a

eIf f a then e x e x if f a

f D f D f a

x e if f af D

x e if f a and so onf a

2. If (x) = cos ax or sin ax and if the differential operator can be written as f(D2) then

)(

12Df

[(x)] = 2

1

( )f a (x) provided f(-a

2) 0.

If f(-a2) = 0 and 2

2 2

1 1f (-a ) 0, then ( (x)) = x. ( (x))

f(D ) f (-a )

25

Example 1.13.8 Solve 2

2

dx

yd - 6

dx

dy + 10 y = cos 2x + e

-3x

Solution: Auxiliary equation is m2 – 6m + 10 = 0.

Therefore m = 1.2

40366 =

6 4

2

=

6 2

2

i = 3 i

Or m = i, where = 3 and = 1.

Therefore C.F. = e3x

(C1 cos x + C2 sin x)

P.I. - 106

12 DD

(cos 2x) + 106

12 DD

(e-3x

)

Using P.I rule 2 and rule 1 respectively we get,

P.I. = 1062

12 D

(cos 2x) + 10)3(6)3(

12

e-3x

= D66

1

(cos 2x) +

10189

1

.e

-3x

= 6

121

1

D

D

(cos 2x) +

37

1e

-3x, multiplying numerator and denominator by 1 + D in the first

expression.

That is, P.I. = 6

1

)2(1

12

D(cos 2x) +

37

1e

-3x

30

1{1 . cos 2x + D(cos 2x)} +

37

1 e

-3x.

That is, P.I = 30

1(cos 2x – 2 sin 2x) +

37

1 e

-3x, since D =

dx

d

General solution is y = C.F. + P.I.

Therefore y = e3x

(C1 cos x + C2 sin x) + 30

1 (cos 2x – 2 sin 2x) +

37

1 e

-3x

Example 1.13.9 Solve (D4 + 18 D

2 + 81) y = cos

3 x.

Solution: Auxiliary equation is m4 + 18 m

2 + 81 = 0

That is (m2 + 9)

2 = 0. Therefore m = 3i, 3i.

Thus C.F. = e0.x

{(C1 + C2 x) cos 3x + (C3 + C4x) sin 3x}

C.F. = (C1 + C2x) cos 3x + (C3 + C4x) sin 3x.

26

P.I = 8118

124 DD

(cos3 x)

= 8118

124 DD

xx 3cos

4

1cos

4

3.

( Since Cos3 A =

4

3 cos A +

4

1 cos 3A).

P.I = 4

3

8118

124 DD

(cos x) + 4

14 2

1

18 81D D (cos 3x)

= 4

3

81)1(18)1(

1222

(cos x) + 4

1

8118

124 DD

(cos 3x)

But f(D2) = D

4 + 18D

2 + 81 and a = 3 in the second expression.

We observe that f(-a2) = f(-9) = 81 – 162 + 81 = 0

Also f 1

(D) = 4D3 + 36 D + 0 = 4 D(D

2 + 9)

Therefore f 1

(-9) = 4 D(-9 + 9) = 0, also

Hence 8118

124 DD

(cos 3x) = x2

)(

1211 Df

(cos 3x)

But f 1

(D) = 4D3 + 36 D, Therefore f

11(D) = 12 D

2 + 36

Hence 8118

124 DD

(cos 3x) = x2,

)3(

1211 f

cos 3x

That is, x2

36)9(12

1

cos 3x = -

72

1 x

2 cos 3x

Thus P.I. = 4

3.

81181

1

cos x +

4

1

xx 3cos

72

1 2

Therefore P.I. = 256

3.cos x -

288

1.x

2 cos 3x.

That is, y = (C1 + C2x) cos 3x + (C3 + C4x) sin 3x + 256

3 cos x -

288

1x

2 cos 3x.

Example 1.13.10 Solve (D2 – 6D + 9)y = x

2 + x + 1

Solution: Auxiliary equation m2 – 6m + 9 = 0.

That is (m – 3)2 = 0. Therefore m = 3, 3

Thus C.F. = (C1 + C2 x)e3x

.

27

P.I = 96

12 DD

(x2 + x + 1)

That is, =

2

9

1

3

219

1

DD

(x2 + x + 1)

= 9

1.

1

2

9

1

3

21

DD (x

2 + x + 1)

(Using the binomial expansion (1 – a)-1

= 1 + a + a2 + …), we have

P.I = 9

1

...

9

1

3

2

9

1

3

21

2

22 DDDD (x2 + x + 1)

= 9

1{1 +

3

2 D -

9

1D

2 +

9

4D

2 + higher powers} (x

2 + x + 1)

= 9

1(1 +

3

2D +

3

1D

2)(x

2 + x+ 1), neglecting D

3 and higher powers since we have only 2

nd degree

polynomial x2 + x + 1.

Therefore P.I. = 9

1{1(x

2 + x + 1) +

3

2

dx

d(x

2 + x + 1) +

3

12

2

dx

d(x

2 + x + 1)}

= 9

1{(x

2 + x + 1) +

3

2(2x + 1) +

3

1(2)}

= 9

1x

2 +

27

7x +

27

7 =

27

1 (3x

2 + 7x + 7)

Thus y = C.F. + P. I. = (C1 + C2x) e3x

+ 27

1(3x

2 + 7x + 7).

Example 1.13.11 Solve 3

3

dx

yd + 3

2

2

dx

yd = 1 + x + e

-3x

Solution: Auxiliary equation is m3 + 3m

2 = 0

Thus C.F. = (C1 + C2x) e0x

+ C3 e-3x

or C1 + C2 x + C3 e-3x

P.I. = 23 3

1

DD (1 + x + e

-3x)

=

313

1

2 DD

(1 + x + e-3x

)

28

= 23

1

D

1

31

D(1 + x) + x .

DD 63

12

e-3x

= 23

1

D

2 3

1 ...3 9 27

D D D

(1 + x) + x.

DD 63

12

e-3x

= 2

1 1 1

3 9 27 81

D

D D

(1 + x) + x.

)3(6)3(3

12

e-3x

= 3

1.

2

1

D(1 + x) -

9

1.

D

1(1 + x) +

27

1(1 + x) -

81

1D(1 + x) +

9

xe

-2x

But D

1(1 + x) = x +

2

2

x,

2

1

D(1 + x) =

2

2x +

6

3x and D(1 + x) = 1.

Therefore P.I. = 3

1

62

32 xx -

9

1

9

2xx +

27

1 (1 + x) -

81

1 (1) +

9

xe

-3x

= 81

2 -

27

2x +

25

162x

2 +

1

18 x

3 +

9

xe

-3x

The term 81

2 -

27

2x may be neglected in view of the arbitrary C1 + C2x appearing in C.F.

General Solution is y = C.F. + P.I. = C1 + C2x + C3 e-3x

+ 25

162x

2 +

1

18 x

3 +

9

xe

-3x

Example 1.13.12 Solve 3

3

dx

yd + 2

2

2

dx

yd +

dx

dy = e

2x + x

3

Solution: Auxiliary equation is m3 + 2m

2 + m = 0

m(m2 + 2m + 1) = 0 or m (m + 1)

2 = 0

Therefore m = 0, -1, -1

Thus C.F. = C1 e0x

+ (C2 + C3 x) e-x

or C1 + (C2 + C3 x) e-x

P.I. = DDD 23 2

1(e

2x) +

DDD 23 2

1(x

3)

Now, DDD 23 2

1(e

2x) =

22.22

123

.e2x

= 18

2xe

Consider DDD 23 2

1(x

3)

Here f(D) = D3 + 2 D

2 + D and f(0) = 0

29

Therefore we write, DDD 23 2

1(x

3) =

)12(

12 DDD

(x3)

= D

1.

2)1(

1

D(x

3) =

D

1(1 + D)

-2 (x

3)

(Using (1 + a)-2

= 1 – 2 a + 3a2 - + …), we have

DDD 23 2

1(x

3) =

D

1(1 – 2D + 3D

2 – 4D

3 + 5 D

4) (x

3), neglecting higher powers of D.

= (D

1 - 2 + 3D – 4D

2 + 5 D

3) (x

3)

Now, D

1 stands for integration, and D, D

2 etc. for successive derivatives.

Therefore DDD 23 2

1(x

3) =

4

4x - 2 .x

3 + 3(3x

2) – 4 (6x) + 5(6).

= 4

4x - 2x

3 + 9x

2 – 24x + 30

Thus P.I. = 18

1 e

2x +

4

4x - 2x

3 + 9x

2 – 24x + 30

The constant term 30 may be neglected in view of the arbitrary constant C1 in the C.F.

Hence, the general solution y = C1 + (C2 e-x

– C3) e-x

+ 18

1e

2x +

4

4x - 2x

3 + 9x

2 – 24x.

Exercise: 1.13.13

1. Solve 4 2

2

dx

yd + 16

dx

dy - 9y = 4 e

x/2 + 3 sin (x/4)

2. Solve (D2 + 1)y = e

x + x

4 + sin x.

Answers :

1. Auxiliary equation is 4m2 + 16 m – 9 = 0. The roots are m =

2

1 and -

2

9 are the roots.

The C.F. = C1 ex/2

+ C2 e-9x/2

P.I. = 9164

12 DD

[4ex/2

+ 3 sin (x/4)]

= 4 9164

12 DD

(ex/2

) + 3. 9164

12 DD

[sin (x/4)].

30

In the first expression f(D) = 4 D2 + 16 D – 9.

Therefore f

2

1 = 4

2

2

1

+ 16

2

1 - 9 = 1 + 8- 9 = 0.

Therefore 9164

12 DD

(ex/2

) = x.168

1

D(e

x/2).

In the second expression, replace D2 by -

24

1.

Therefore P.I. = 4x 168

1

D(e

x/2) + 3.

9164

14

1

2

D

[sin (x/4)]

P.I = 4x.

162

18

1

.e

x/2 + 3.

9164

1

1

D

[sin (x/4)]

= 5

1xe

x/2 + 3.

2

2

4

37256

4

3716

D

D

[sin (x/4)]

Thus P.I = 5

1xe

x/2 +

2

2 4

37

4

1256

4

37163

D

[sin (x/4)]

= 5

1 xe

x/2 -

1625

12(64 D + 37) sin (x/4)

General solution is

y = C.F + P.I. = C1 ex/2

+ C2 e-9x/2

+ 5

1xe

x/2 -

1625

12{16 cos (x/4) + 37 sin (x/4)}

2. The roots of the auxiliary equation is m = i.

Therefore C.F. = e0x

(C1 cos x + C2 sin x) that is, = C1 cos x + C2 sin x.

P.I = 1

12 D

(ex + x

4 + sin x) =

2

1e

x + (x

4 – 12x

2 +24) + x .

2

1(-cos x )

General solution is y = C1 cos x + C2 sin x + 2

1 e

x + x

4 –12 x

2 + 24 -

2

x cos x.

1.14 Variation of Parameters

31

Consider the second order linear differential equation

( ) ( ) ( ). (1)y p x y q x y R x

Suppose that 1 1 2 2cy c y c y is a solution, where

1 2 and y y are linearly independent on an

interval .a x b Let us see what happens if we replace both of the constants 1 2 c and c with

functions of x.

We consider 1 2 (2)y Ay By and try to determine ( ) ( )A x and B x so that

1 2Ay By is a solution of the equation (1).

Then 1 2 1 2. (3)y Ay By A y B y

Rather than involved with the derivatives of ( ) ( )A x and B x of higher order than the first, we

now choose some particular function for the expression 1 2A y B y .

For simplicity we choose 1 2 0. (4)A y B y

It then follows from (3) that 1 2 1 2 . (5)y Ay By A y B y

Since y was to be a solution of (1) we substitute from (2), (3), and (5) into equation (1) to obtain

1 1 1 2 2 2 1 2( ) ( ) ( ).A y py qy B y py qy A y B y R x

But 1 2 and y y are solutions of the homogeneous equation, so that finally

1 2 ( ). (6)A y B y R x

Equations (4) and (6) now give us two equations that we wish to solve for ( ) ( ).A x and B x

This solution exists providing the determinant 1 2

1 2

( )y y

W xy y

does not vanish. But this

determinant is precisely the Wronskian of the functions1 2 and y y , which are presumed to be

linearly independent on the interval .a x b Therefore, the Wronskian does not vanish on that

interval and we can find .A and B By integration we get

2 1( ) ( )( ) ( )

( ) ( )

y R x y R xA x dx and B x dx

W x W x

Example 1.14.1 Solve the equation 2( 1) sec tan .D y x x

Immediately we find that 1 2cos sin .cy c x c x

32

Let us seek a particular solution by variation of parameters. Put cos sin .py A x B x

Then

1 2

1 2

cos sin( ) 1.

sin cos

y y x xW x

x xy y

22 ( ) sin sec tan( ) tan tan .

( ) 1

y R x x x xA x dx dx xdx x x

W x

(Constant of integration has been disregarded because we are seeking only a particular solution.)

And

1 ( ) cos sec tan( ) tan ln | sec | .

( ) 1

y R x x x xB x dx dx xdx x

W x

Therefore, the complete solution is

1 2

1 3

( ) cos sin cos ( tan ) sin ln | sec |

cos sin cos sin ln | sec |,

y x c x c x x x x x x

c x c x x x x x

where the term ( sin )x in py has been absorbed in the complementary function term

3 sinc x ,

since 3c is an arbitrary constant.

Example 1.14.2 Solve 2 1

( 3 2) .1 x

D D ye

Here 2

1 2 ,x x

cy c e c e so we put

2 .x x

py Ae Be

Then

21 2 3

2

1 2

( ) .2

x x

x

x x

y y e eW x e

e ey y

2

2

-x 3x

( )( ) ln(1 ).

( ) (1+e )e 1

x xx

x

y R x e eA x dx dx dx e

W x e

And

2

1 ( )( ) ln(1 ).

( ) (1 ) 1

x xx x x

x x

y R x e eB x dx dx e dx e e

W x e e

33

Therefore, the complete solution is 2 2

3 2( ) ( )ln(1 ).x x x x xy x c e c e e e e

Exercises 1.14.3

Solve using variation of parameters:

1. 2( 1) csc cot .D y x x

2. 2 4( 1) sec .D y x

3. 2 2( 1) tan .D y x

4. 2 2( 1) sec csc .D y x x

5. 2 2 2( 2 1) ( 1) .x xD D y e e

6. 2( 3 2) cos( ).xD D y e

7. 2 2 1/ 2( 1) 2(1 ) .xD y e

8. 2 2( 1) sin .x xD y e e

1.14 Cauchy’s homogeneous linear equation:

This equation is of the form

1

1

1 11... ................ 1

n nn n

n nn n

d y d y dyx k x k x k y X

dx dx dx

, where X is a function of x , and ik ,

1,2..., ,i n are constants. Equations of this type can be reduced to linear differential equations

with constant coefficients by letting tx e . Thus logt x .

If ,d

Ddt

then 1

. .dy dy dt dy

dx dt dx dt x ; i.e.,

dyx Dy

dx

2 2

2 2 2

1 1.

d y d dy d y dy

dx dx x dt x dt dt

; i.e.,

22

21

d yx D D y

dx .

Similarly, 3

3

31 2

d yx D D D y

dx , and so on.

After making these substitutions in equation (1), we get a linear equation with constant

coefficients which can be solved as before.

Example 1.15.1 Solve 2

2 2

24 6

d y dyx x y x

dx dx

Solution: Put tx e . Then logt x . Let ,d

Ddt

then dy

x Dydx

, 2

2

21

d yx D D y

dx .

The given equation becomes 21 4 6 tD D D y e ; i.e., 2 25 6 tD D y e

The auxiliary equation is 2 5 6 0m m ; i.e., 2 3 0m m .

34

The roots are 2,3m . The complementary function is 2 3

1 2

t t

cy c e c e

The particular integral is 2

2

1

5 6

t

py eD D

21

2 5

tt eD

2

2 2 5

tet

2tte

The complete solution is 2 3 2

1 2 logc py y y c x c x x x

Example 1.15.2 Solve 2

2 3

22 12 log

d y dyx x y x x

dx dx

Solution: Put tx e . Then logt x . Let ,d

Ddt

then dy

x Dydx

, 2

2

21

d yx D D y

dx .

The given equation becomes 2 312 tD D y te

The roots of the auxiliary equation are 3, 4m

The complementary function is 3 4

1 2

t t

cy c e c e

The particular integral is 3

2

1

12

t

py teD D

3 2 1 1

7 2 7 49

te tt

The complete solution is

233 4

1 2

log 1 1log

7 2 7 49c p

xxy y y c x c x x

Example 1.15.3 Solve 3 2

3 2

3 23 log

d y d y dyx x x y x x

dx dx dx

Solution: Put tx e . Then logt x .

Let ,d

Ddt

then dy

x Dydx

, 2

2

21

d yx D D y

dx ,

33

31 2

d yx D D D y

dx

The given equation becomes 3 1 tD y e t

The roots of the auxiliary equation are 1 3

1,2

im

;

The complementary function is 21 2 3

3 3cos sin

2 2

tt

cy c e e c t c t

The particular integral is 3

1

1

t

py e tD

2

tet

The complete solution is

1

1 2 3

3 3cos log sin log log

2 2 2c p

xy y y c x x c x c x x

Exercise 1.15.4

Solve

1. 2

2 4

22 4

d y dyx x y x

dx dx

35

2. 2

2 2

2 1d yx y x

dx x x

3. 2

2

2sin log log

d y dyx x y x x

dx dx

Answers

1. 4

1 4

1 2 log5

c p

xy y y c x c x x

2. 2 2

1 2

1 log 1

3

xy c x c x

x x

3. 2

1 2

1 1cos log sin log log cos log log log sin log

4 4c py y y c x c x x x x x

1.16 Legendre’s linear equation:

This equation is of the form

1

1

1 11... ................ 1

n nn n

n nn n

d y d y dyax b k ax b k ax b k y X

dx dx dx

,

where X is a function of x , and ik , 1,2..., ,i n are constants. Equations of this type can be

reduced to linear differential equations with constant coefficients by letting tax b e . Thus

logt ax b .

If ,d

Ddt

then 1

. . .dy dy dt dy

adx dt dx dt ax b

; i.e., dy

ax b aDydx

2 2

221

d y aD D y

dx ax b

; i.e.,

22 2

21

d yax b a D D y

dx .

Similarly, 3

3 3

31 2

d yax b a D D D y

dx , and so on.

After making these substitutions in equation (1), we get a linear equation with constant

coefficients which can be solved as before.

Example 1.16.1 Solve 2

2

22 3 2 3 12 6

d y dyx x y x

dx dx

Solution: Put 2 3 tx e . Then log 2 3t x .

Let ,d

Ddt

then 2 3 2dy

x Dydx

, 2

2 2

22 3 2 1

d yx D D y

dx .

The given equation becomes 3

4 1 2 12 62

teD D D y

;

i.e., 24 6 12 3 9tD D y e

The roots of the auxiliary equation are 3 57

4m

;

The complementary function is 3 57 3 57

4 41 2

t t

cy c e c e

36

The particular integral is 2

13 9

4 6 12

t

py eD D

3 3

14 4

te

The complete solution is 3 57 3 57

4 41 2

3 32 3 2 3 2 3

14 4c py y y c x c x x

Example 1.16.2 Solve 3 2

3 2

3 21 2 1 4 1 4 4log 1

d y d y dyx x x y x

dx dx dx

Solution: Put 1 tx e . Then log 1t x . Let ,d

Ddt

then 1dy

x Dydx

,

2

2

21 1

d yx D D y

dx ,

33

31 1 2

d yx D D D y

dx

The given equation becomes 3 2 4 4 4D D D y t

The roots of the auxiliary equation are 1,2, 2m ;

The complementary function is 2 2

1 2 3

t t t

cy c e c e c e

The particular integral is 3 2

14

4 4py t

D D D

12

1 ( ...) ( ) 14

DD t t

The complete solution is 2 2

1 2 31 1 1 log 1 1c py y y c x c x c x x

Example 1.16.3 Solve 2

2

21 1 sin 2log 1

d y dyx x y x

dx dx

Solution: Put 1 tx e . Then log 1t x .

Let ,d

Ddt

then 1dy

x Dydx

, 2

2

21 1

d yx D D y

dx .

The given equation becomes 2 1 sin 2D y t

The roots of the auxiliary equation are m i ;

The complementary function is 1 2cos sincy c t c t

The particular integral is 2

1sin 2

1py t

D

1sin 2

3t

The complete solution is

1 2

1cos log 1 sin log 1 sin 2log 1

3c py y y c x c x x

Exercise 1.16.4

Solve

1. 2

2 2

23 2 5 3 2 3 1

d y dyx x y x x

dx dx

2. 2

2

21 1 2sin log 1

d y dyx x y x

dx dx

3. 2

2 2

22 1 2 1 2 8 2 3

d y dyx x y x x

dx dx

Answers

37

1. 1 1 2

31 2

1 7 13 2 3 2 3 2 3 2

405 27 108y c x c x x x

2. 1 2coslog 1 sin log 1 log 1 cos log 1y c x c x x x

3. 1 2

21 2

1 12 1 2 1 2 1 2 1 log 2 1 2

5 2y c x c x x x x

1.17 System of linear differential equations with constant coefficients

Quite often we come across a system of linear differential equations with constant coefficients in

which there are two or more dependent variables and a single independent variable exists. Such

a system of equation can be solved by eliminating all but one of the dependent variables , and

solving the resulting equation. Then using the given equations, the other dependent variables

can be expressed in terms of the dependent variable which is obtained earlier and can be

determined.

Example 1.17.1 Solve the simultaneous equations:

5 2 ; 2 0dx dy

x y t x ydt dt

being given 0x y when 0.t

Solution: Taking ,d

Ddt

the given equations become

( 5) 2 ( )

2 ( 1) 0 ( )

D x y t i

x D y ii

Eliminate x as if D were an ordinary algebraic multiplier. Multiplying (i) by 2 and operating on

(ii) by (D+5) and then subtracting, we get

2( 6 9) 2 .D D y t

Its complementary function is 3

1 2( ) ( ) t

cy t c c t e and a particular integral is

2

1 2 4( ) ( 2 ) .

( 3) 9 27p

ty t t

D

Thus ( ) .c py t y y

When t=0, 1

40 .

27y c

Substituting the value of y in (ii), we obtain

3

2 2

4 1 1( ) .

27 2 9 27

t tx t c c t e

When t = 0, 2

20 .

9x c

Hence the desired solutions are

3 31 1 2 2( ) (1 6 ) (1 3 ); ( ) (2 3 ) (2 3 ).

27 27 27 27

t tx t t e t y t t e t

38

Example 1.17.2 Solve the simultaneous equations

2 sin 0; 2 cos 0dx dy

y t x tdt dt

given that x = 0 and y = 1 when t = 0.

Solution: Taking ,d

Ddt

the given equations become

2 sin ( )

2 cos ( )

Dx y t i

x Dy t ii

Eliminating x by multiplying (i) by 2 and operating on (ii) by D and then adding, we get

2( 4) 3sin .D y t

Its complementary function is 1 2( ) cos2 sin2cy t c t c t and a particular integral is

2

1( ) 3 sin sin .

4py t t t

D

Thus ( ) .c py t y y

When t=0, 11 1.y c

Substituting the value of y in (ii), we obtain

2( ) sin2 cos2 cos .x t t c t t

When t = 0, 20 1.x c

Hence the desired solutions are

( ) cos2 sin2 cos ; ( ) cos2 sin2 sin .x t t t t y t t t t

Example 3:

Solve the simultaneous equations

2 2cos 7sin ; 2 4cos 3sin .dx dy dx dy

y t t x t tdt dt dt dt

Solution: Taking ,d

Ddt

the given equations become

( 2) 2cos 7sin ( )

( 2) 4cos 3sin ( )

Dx D y t t i

D x Dy t t ii

Eliminating y by operating on (i) by D and operating on (ii) by (D-2) and then adding, we get 2( 2) 9cos .D x t

Its complementary function is 2 2

1 2( ) t t

cx t c e c e and a particular integral is

( ) 3cos .px t t

Thus ( ) .c px t x x

Substituting the value of x in (ii), we obtain

39

2 2

1 2 3( ) ( 2 1) ( 2 1) 2sin .t tx t c e c e t c

Where 1 2 3, and c c c are arbitrary constants.

Exercise:

Solve

i. sin 0; cos 0dx dy

y t x tdt dt

given that x = 2 and y = 0 when t = 0.

ii. ( 1) 2 1;(2 1) 2D x Dy t D x Dy t .

iii. ( 1) (2 1) ;( 1) ( 1) 1tD x D y e D x D y .

iv. 2( 1) 4( 1) 4 ;( 1) ( 9) 0xD y D v e D y D v given y = 5, dy/dx = 0, v=1/2

at x = 0. ( Hint: Eliminate v(x) first and solve for y(x))

2. The small oscillations of a certain system with two degrees of freedom are given by the

equations 2 2 2( 3) 2 0;( 3) ( 5) 0D x y D x D y where

dD

dt . If x = y =

0, Dx=3, Dy = 2 when t = 0, find x and y when t = ½.

3. A mechanical system with two degrees of freedom satisfies the equations 2 2

2 22 3 4;2 3 0.

d x dy d y dx

dt dt dt dt Obtain expressions for x and y in terms of t, given

x, y, dx/dt, dy/dt all vanish at t = 0.