Linear ODE

HOMOGENEOUS

CONSTANT COEFFICIENT

VARIABLE COEFFICIENT

SEULER-CAUCHY

NON-HOMOGENEO

US

CONSTANT COEFFICIENT

VARIABLE COEFFICIENT

SEULER-CAUCHY

NON-HOMOGENEOUS O.D.E. WITH CONSTANT COEFFICIENTS

Solution of Non-Homogeneous ODE

Solution of Corresponding Homogeneous ODE(C.F.)

Particular Integral Of Given Non-Homogeneous ODE(P.I)

General Solution =C.F.+P.I.

Method of Finding Particular

Integral

Method of Undetermined Coefficients

Method of Variation of Parameters

1

F(D)

Method of Undetermined Coefficients

The Following Table shows the choice of

The Choice for Yp is made using the following three rules on the basis of the table as well.

r(x)

Basic Rule If r(x) is one of the functions given in the first column of

the table 2.1, choose Yp in the

same line.

Modification Rule If a term in your choice of Yp happens to be a basic solution of the Homogeneous ODE then multiply your

choice by x

Sum Rule If r(x) is sum of

functions in the first column of the Table 2.1, choose for Yp the sum of the functions in the corresponding lines of the second column.

r(x) Basic Rule

The Application of Basic Rule

Solve2y''+ y = x

2

2

0x

h

h

(D +1)y = 0

AuxiliaryEquation :D +1= 0

D = ±i

C.F. y = e (A cosx+Bsinx)

y (C.F.) = A cosx+Bsinx

=

Step-1: General Solution of the homogeneous ODE

Step-2: Solution Yp of the Non-homogeneous ODE

We look for choice Yp from the table for the given 2 nr(x) = x (Of theformx ,n )= 2

From the following table

2p 0 1 2

''p

2

2 22 0 1 2

Let y = K + K x+ K x .

Theny = 2K .

Bysubstituting this iny''+ y = x weget,

2K + K + K x+ K x = x x x= + + 20 0 1

2 0

1

2 02

p

By comparing the coefficients,

(2K + K ) = 0

K = 0

K =1 K = -2

So,y (P.I.) = -2+ x is the required P .I .

Þ

GENERAL SOLUTION(G.S.=C.F.+P.I.)

y(x) A cosx Bsinx x .= + - + 22

h py(x) y (x) y (x)= +

r(x) Modification Rule

The Application of Modification Rule

xSolve: y'' y' y e-+ + = 23 2 30

2

2

x xh

x x

(D + D )y = 0

AuxiliaryEquation :D + D = 0

(D )(D )

D ,

C.F. y = c e c e

y e and y e

- -

- -

+

+

Þ + + =

Þ = - -

= +

= =

21 2

21 2

3 2

3 2

1 2 0

1 2

Step-1: General Solution of the homogeneous ODE

Solution of homogeneous equation : y'' y' y+ + =3 2 0

Step-2: Solution Yp of the Non-homogeneous ODE

We look for choice Yp from the table for the given 2x xr(x) = e (Of theformKe )-30

From the following table

xy e-= 22

' x xp

'' x xp

'' ' xp p p

x x x x x x

x x x x x x

Theny Ke Kxe

&y Ke Kxe

Substitutingthesevaluesininthegiven diff.eqn.

y y y e

Ke Kxe (Ke Kxe ) Kxe e

Ke Kxe Ke Kxe Kxe e

- -

- -

-

- - - - - -

- - - - - -

= -

= - +

+ + =

Þ - + + - + =

Þ - + + - + =

2 2

2 2

2

2 2 2 2 2 2

2 2 2 2 2 2

2

4 4

3 2 30

4 4 3 2 2 30

4 4 3 6 2 30x x

xp

Ke e

By comparing the coefficient

K

y (P.I.) = xe is the required P .I .

- -

-

Þ - =

= -

-

2 2

2

30

30

30

But if we look at the C.F. it has the solution

which is same as choice of x

py Ke-= 2

So, by modification rule we have to multiply the above choice of Yp

by x, i.e., we have the modified choice xp

y Kxe-= 2

GENERAL SOLUTION(G.S.=C.F.+P.I.)

x x xy(x) c e c e xe .- - -= + -2 21 2

30

h py(x) y (x) y (x)= +

r(x) Sum Rule

The Application of Sum Rule

Solve: y'' y' y x sin x+ + = +24 5 25 13 2

x

hx

h

(D D )y = 0

AuxiliaryEquation :

D D = 0

D i,

C.F. y = e (A cosx+Bsinx)

y (C.F.) = e (A cosx+Bsinx)

-

-

+ +

+ +

- + -= = - ±

=

2

2

2

2

4 5

4 5

4 16 202

2

Step-1: General Solution of the homogeneous ODE

Solution of homogeneous equation : y'' y' y+ + =4 5 0

Step-2: Solution Yp of the Non-homogeneous ODE

We look for choice Yp from the table for the given r(x) = x sin x r (x) r (x)+ = +2

1 225 13 2

From the following table

n(Here r (x)is of theformx ,n

& r (x) is of the formksin x, ) =

=1

2

2

2

2p 0 1 2

'p''p

Let y = (K + K x+ K x ) (K sin x K cos x).

Theny = K K x K cos x K sin x.

y = K K sin x K cos x.

+ +

+ + -

- -

3 4

1 2 3 4

2 3 4

2 2

2 2 2 2 2

2 4 2 4 22

20 1 2

Bysubstituting this iny''+ y' y = x sin x,weget,

( K K sin x K cos x)

(K K x K cos x K sin x)

(K + K x+ K x K sin x K cos x)

x x sin x cos x

+ +

- -

+ + + -

+ + +

= + + + +

2 3 4

1 2 3 4

3 42

4 5 25 13 2

2 4 2 4 2

4 2 2 2 2 2

5 2 2

0 0 25 13 2 0 2

( K K K ) ( K K )x K x

( K K K )sin x ( K K K )cos x

x sin x

Þ + + + - - +

+ - - + + - + +

= +

22 1 0 2 1 2

3 4 3 4 3 42

2 4 5 8 5 5

4 8 5 2 4 8 5 2

25 13 2

( K K K ) ( K K )x K x

(K K )sin x (K K )cos x x sin x

Þ + + + - - +

+ - + + = +

22 1 0 2 1 2

23 4 4 3

2 4 5 8 5 5

8 2 8 2 25 13 2

( K K K ) ( K K )x K x

(K K )sin x (K K )cos x x sin x

Þ + + + - - +

+ - + + = +

22 1 0 2 1 2

23 4 4 3

2 4 5 8 5 5

8 2 8 2 25 13 2

By comparing the coefficients we get,

K K K , K K , K ,

K K , K K

+ + = - - = =

- = + =2 1 0 2 1 2

3 4 4 3

2 4 5 0 8 5 0 5

8 13 8 0

KK K &K K .

& K K , K K ( i.e.,K K )

Þ - - = = Þ = - = -

- = + = = -

22 1 2 1

3 4 4 3 4 3

88 5 0 5 8

58 13 8 0 8

K ( K ) gives K K

K K &K K .

using these in K K K ,

we have K K

Þ - - = + =

Þ = Þ = = - = -

+ + =

- + = Þ =

3 3 3 3

3 3 4 3

2 1 0

0 0

8 8 13 64 13

1 865 13 8

5 52 4 5 0

2210 32 5 0

5

pSo, y (P.I.) x x (sin x cos x)= - + + -222 1

8 5 2 8 25 5

GENERAL SOLUTION(G.S.=C.F.+P.I.)

xy(x) e (A cosx+Bsinx) x x (sin x cos x).

-= + - + + -

2 222 18 5 2 8 2

5 5

h py(x) y (x) y (x)= +

4251 3

0011 0010 1010 1101 0001 0100 1011Method of Finding Particular Integral

Method of Variation of Parameters

4251 3

0011 0010 1010 1101 0001 0100 1011

Method of Variation of Parameters

4251 3

0011 0010 1010 1101 0001 0100 1011

x

h

h

(D )y = 0

AuxiliaryEquation :D = 0 D i,

C.F. y = e (A cosx+Bsinx)

y (C.F.) = A cosx+Bsinx

Next,we find theP.I.

Here we have y cosx and y sinx

+

+ Þ = ±

=

= =

2

2

0

1 2

1

1

Solution of homogeneous equation : y'' y+ = 0

' 'So,W W(y ,y ) y y y y

cosx(cosx) sinx( sinx) cos x sin x

= = -

= - - = +

=

1 2 1 2 2 12 2

1

4251 3

0011 0010 1010 1101 0001 0100 1011

Next, we have the formula,

General solution is

Method of variation of parameres for higher order ODE