Linear differential equation with constant coefficient

Linear Differential Equation
with constant coefficient
Sanjay Singh
Research Scholar
UPTU, Lucknow

The order linear differential equation
with constant coefficient
th
n
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 dy
a a a a a y Q
dxdx dx dx
− −
−− −
+ + + + + =
3 2
3 2
3 6 2 sin 5
Example
d y d y dy
y x
dxdx dx
+ − + =

( )F D y Q=
If
d D
dx
=
1 2
1 2 1( ) .......n n n
o n nWhere F D a D a D a D a D a− −
−= + + + + +
3 2
3 2
3 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 n
OR a m a m a m a m a− −
−
+ + + + + =
3 2
3 2
3 6 2 sin5
d y d y dy
y x
dxdx dx
Example + − + =
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 2
1 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 2
1 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 n
o 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 2
1 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 4
1 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= = =
542
1 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 xx
nC 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= = + = = −
5
1 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 4
1 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
, ,......, n
andm m i m m i m mα β α β= = + = = −
3
1 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
: ax
when QCASE I e=
1 1
. , ( ) 0
( ) ( )
ax axP I F a
F D F a
e e= ≠=
( ) 0F a =
'
'
1 1
. , ( ) 0
( ) ( )
ax axP I F a
F D F a
e x e= ≠=

# if then
Case II: when
'
( ) 0F a =
2
'
''
''
'
1
. .
( )
1
,
( )
1
,
( )
( ) 0
( ) 0
0
,
( )
ax
ax
ax
then P I
F D
x
F D
F
F a
x
F a
F a
e
e
e a
=
=
=
=
=
≠
sin cos( )Q ax or ax b= +
2 2
2 2
1
. ( )
[ ( )]
1
( ), [ ( )] 0
[ ( )] D a
D a
P I Sin ax b
F D
Sin ax b F D
F D =−
=−
= +
= + ≠

# if
# if
2 2
'
[ ( )] 0D a
If F D = −
=W
2 2
2 2
2 2
'
'
1
. ( ), [ ( )] 0
[ ( )]
1
( ), [ ( )] 0
[ ( )]
D a
D a
D a
P I Sin ax b F D
F D
x Sin ax b F D
F D
=−
= −
=−
= + =
= + ≠
2 2
'
[ ( )] 0D a
F D = −
=
2
2 2
2 2
2 2
2 2
'
'
''
''
1
. ( ), [ ( )] 0
[ ( )]
1
( ), [ ( )] 0
[ ( )]
1
( ), [ ( )] 0
[ ( )]
D a
D a
D a
D a
P I Sin ax b F D
F D
x Sin ax b F D
F D
x Sin ax b F D
F D
= −
= −
= −
= −
= + =
= + =
= + ≠

Case III: when , m non negative integer
Expending by Binomial theorem P.I. can be
evaluated
m
Q x=
1
1
.
( )
1
deg [1 ( )]
1
[1 ( )] ( )
m
m
m
P I x
F D
x
Lowest ree term D
D x
LDT
φ
φ −
=
±
= ±
=
1
[1 ( )]Dφ −
±

Case IV: when
Case V: (General Method), Q is any function of ‘x’
ax
Q e V=
1 1
.
( ) ( )
ax ax
P I e V e V
F D F D a
= =
+
1 1
.
( ) ( )( )
1 1
( ) ( )
1
( )
x x
P I Q Q
F D D D
Q
D D
Q dx
D
e eα α
φ α
φ α
φ
−
= =
−
 
=  − 
= ∫

1. Solve
Solution: The d.e. is
The A.E. is
Factorizing
The roots are
2 2
3 2 2
2 2
2 2
1 1
. .
( 3 4) 3 6
1
.
(6 6) 6
x x
x
x
P I e x e
D D D D
x e
x e
D
= =
− + −
= =
−
The complete solution is

2. Solve
The a.e. is
Factorizing
The roots are

3. Solve
The a.e. is Factorizing
The roots are
And

4. Solve
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 '
1sin2 cos2 2 ( cos sin ) co
2
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 isgivenby
= + − −−−−−−−−−
⇒ − + = +
⇒ = + + −
= + + + −
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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