linear differential equation with constant coefficient

Linear Differential Equation with constant coefficient

Sanjay SinghResearch Scholar UPTU, Lucknow

The order linear differential equation with constant coefficient

thn

1 2

0 1 2 11 2.......

n n n

n nn n n

The Differential Equation of the form

d y d y d y dya a a a a y Q

dxdx dx dx

− −

−− −+ + + + + =

3 2

3 23 6 2 sin 5

Example

d y d y dyy x

dxdx dx+ − + =

( )F D y Q=

If d Ddx

=

1 21 2 1( ) .......n n n

o n nWhere F D a D a D a D a D a− −−= + + + + +

3 2

3 23 6 2 sin5Example d y d y dy y x

dxdx dx+ − + =

3 2( 3 6 2 ) 5D y D y Dy y Sin x⇒ + − + =3 2( 3 6 2) 5D D D y Sin x⇒ + − + =( ) 5F D y Sin x⇒ =

3 2( ) ( 3 6 2)F D D D D∴ = + − +

Auxiliary Equation(A.E.)

. . . ( )Suppose L D E is F D y Q=. . ( ) 0A E is F m =

1 2

1 2 1....... 0n n n

o n nOR a m a m a m a m a− −−+ + + + + =

3 2

3 23 6 2 sin5

d y d y dyy x

dxdx dxExample + − + =

3 2( 3 6 2) 5D D D y Sin x⇒ + − + = ( ) 5F D y Sin x⇒ =3 2( ) ( 3 6 2)F D D D D∴ = + − +

3 2. . ( ) 0 3 6 2 0Hence A E is F m m m m= ⇒ + − + =

Complementary Function (C.F.) of L.D.E. A function of ‘x’ which satisfies the L.D.E is known as complementary function of L.D.E . .

Particular Integral (P.I.) of L.D.E.A function of ‘x’ which satisfies the L.D.E. is known as particular integral of L.D.E .

General Solution of L.D.E. The general solution of L.D.E is given by

y = C.F. + P.I

( ) 0F D y =

( )F D y Q=

( )F D y Q=

General Solution of L.D.E.

Complete Solution :

y = C.F+P.I Where C.F Complementary Function

P.I Particular Integral

. . . ( )Suppose L D E is F D y Q=

Complementary Function

A function of ‘x’ which satisfies the L.D.E

F(D)y = 0

is known as complementary function of L.D.E . .

Determination of C.F. Consider the L.D.E . F(D)y = 0

Write A.E. of L.D.E. F(m) = 0

Solve A.E.

Suppose are the ‘n’ roots of the auxiliary equation.

1 21 2 1....... 0n n n

o n na m a m a m a m a− −−⇒ + + + + + =

1 2 3, , ,........., nm m m m

Case I: (Roots are real)1 2 3, , ,........., nIf m m m m are distinctW

31 21 2 3. ....... nm xm x m x m x

nthen C F c e c e c e c e= + + + +

Determination of C.F.

Consider the L.D.E .

# Write A.E. of L.D.E.

i.e. .

# Solve A.E.Suppose are the ‘n’ roots of the auxiliary equation.

# Case I: (Roots are real)

# If are distinct then

( )F D y Q=

( ) 0F m =1 2

1 2 1....... 0n n no n na m a m a m a m a− −

−+ + + + + =

1 2 3, , ,........., nm m m m

1 2 3, , ,........., nIf m m m m are distinct

31 21 2 3. ....... nm x m xm x m x

nthen C F c e c e c e c e= + + + +





1 2 3 4,( ) , ........., nm m k say and m m m= =

3 41 2 3 4. ( ) ....... nm x m xm xkx

nC F c c x e c e c e c e= + + + +

1 2 3 4 5,( ) , ........., nm m m k say and m m m= = =

5421 2 3 4 5. ( ) ....... nm x m xm xkx

nC F c c x c x e c e c e c e= + + + + +

1 2 3 4,, ........., nm m and m m mα β α β= + = −3 4

1 2 3 4. ( cosh sinh ) ....... nm x m xm xxnC F e c x c x c e c e c eα β β= + + + +

1 2 3 4 5, , ,......, nm m m m and m mα β α β= = + = = −

51 2 3 4 5. [( ) cosh ( )sinh )] ...... nm x m xx

nC F e c c x x c c x x c e c eα β β= + + + + + +

# Case II: (Roots are comlex)

# If are real and distinct then

# If are real and distinct then

1 2 3 4, , ......, nm i m i and m m mα β α β= + = −

3 41 2 3 4. ( cos sin 0 ....... nm x m xm xx

nC F e c x c x c e c e c eα β β= + + + +

1 2 3 4 5, , ......, nandm m i m m i m mα β α β= = + = = −

31 2 3 4 5. [( ) cos ( )sin ] ..... nm x m xx

nC F e c xc x c xc x c e c eα β β= + + + + + +

Determination of P.I.Determination of P.I.

P.I. of L.D.E. is given by

Thus P.I. =

Case I: when

# If then

( )F D y Q= 1

( )Q

F D

1

( )Q

F D

: axwhen QCASE I e=1 1

. , ( ) 0( ) ( )

ax axP I F aF D F a

e e= ≠=

( ) 0F a =

''

1 1. , ( ) 0

( ) ( )ax axP I F a

F D F ae x e= ≠=

# if then

Case II: when

'( ) 0F a =

2

'

''''

'

1. .

( )

1,

( )

1,

( )

( ) 0

( ) 0

0

,

( )

ax

ax

ax

then P IF D

xF D

FF a

x

F a

F a

e

e

e a

=

=

=

=

=

≠

sin cos( )Q ax or ax b= +

2 22 2

1. ( )

[ ( )]

1( ), [ ( )] 0

[ ( )] D aD a

P I Sin ax bF D

Sin ax b F DF D =−

=−

= +

= + ≠

# if

# if

2 2'[ ( )] 0

D aIf F D = − =W

2 2

2 2

2 2

''

1. ( ), [ ( )] 0

[ ( )]

1( ), [ ( )] 0

[ ( )]

D a

D aD a

P I Sin ax b F DF D

x Sin ax b F DF D

=−

= −=−

= + =

= + ≠

2 2'[ ( )] 0

D aF D = − =

2

2 2

2 2

2 2

2 2

''

' '' '

1. ( ), [ ( )] 0

[ ( )]

1( ), [ ( )] 0

[ ( )]

1( ), [ ( )] 0

[ ( )]

D a

D a

D aD a

P I Sin ax b F DF D

x Sin ax b F DF D

x Sin ax b F DF D

= −

= −

= −= −

= + =

= + =

= + ≠

Case III: when , m non negative integer

Expending by Binomial theorem P.I. can be evaluated

mQ x=

1

1.

( )

1

deg [1 ( )]

1[1 ( )] ( )

m

m

m

P I xF D

xLowest ree term D

D xLDT

φ

φ −

=

±

= ±

=

1[1 ( )]Dφ −±

Case IV: when

Case V: (General Method), Q is any function of ‘x’

axQ e V=

1 1.

( ) ( )ax axP I e V e V

F D F D a= =

+

1 1.

( ) ( )( )

1 1

( ) ( )

1

( )x x

P I Q QF D D D

QD D

Q dxD

e eα α

φ α

φ α

φ−

= =−

= −

= ∫

1. Solve

Solution: The d.e. is

The A.E. is

Factorizing The roots are

2 23 2 2

2 22 2

1 1. .

( 3 4) 3 6

1 .

(6 6) 6

x x

xx

P I e x eD D D D

x ex e

D

= =− + −

= =−

The complete solution is

2. Solve

Solution: The d.e. is The a.e. is

Factorizing

The roots are


3. Solve


The a.e. is Factorizing

The roots are

And


4. Solve

Solution: The d.e. is The a.e. is


∴

∴

5. Solve


The a.e. is

Factorizing

The roots of A.E. are


6. Solve Solution:

Here

But ,

and


Legendre’s Linear Equations A Legendre’s linear differential equation is of the form

where are constants and This differential equation can be converted into L.D.E with constant coefficient by subsitution

and so on

Note: If then Legendre’s equation is known as Cauchy- Euler’s equation

7. Solve

Put Then

The C.S. is

Simultaneous Linear Differential Equations

The most general form a system of simultaneous linear differential equations containing two dependent variable x, y and the only independent variable t is

…………………(1),

where are constants and and are functions of t only.

8. Solve :

Solution: The system is

Eleminating ‘y’ between Equations (1) and (2), we get

It is L.D.E. with constant coefficient.

1 2

1 2

1( cos sin ) cos2 . (4)2

(1)+(2) 2 ' 2 2 sin2 cos2

2 sin2 cos2 2 2 '

1 sin2 cos2 2 ( cos sin ) co2

From (1) and (2),

(3)

t

t

x e C t C t t

x x y t t

y t t x x

t t e C t C t

Solution of eqn is givenby

= + − − − − − − − − − −

⇒ − + = +⇒ = + + −

= + + + −

1 2 1 2

1 2 1 2 1 2

s2

2 ( cos sin ) ( sin cos ) sin2 by using (3)

2 [ cos sin cos sin sin cos ]

sin2 cos2

t t

t

t

e C t C t e C t C t t

e C t C t C t C t C t C t

t

− + + − + +

= + − − + −

+ +

1 2

1 2

cos2 2sin2

2 ( sin cos ) sin2

1( sin cos ) sin2 ....................(5)2

(5) (6) tan .

t

t

t t t

e C t C t t

C t C t ty e

Equations and give complete solution of given simul eous equations

− −= − −

− −∴ =

linear differential equation with constant coefficient

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