linear differential equation with constant coefficient
TRANSCRIPT
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= + + + +
# If are distinct then
# If are distinct then
# If are distinct then
# If are distinct then
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
The complete solution is
3. Solve
Solution: The d.e. is
The a.e. is Factorizing
The roots are
And
The complete solution is
4. Solve
Solution: The d.e. is The a.e. is
The complete solution is
∴
∴
5. Solve
Solution: The d.e. is
The a.e. is
Factorizing
The roots of A.E. are
The complete solution is
6. Solve Solution:
Here
But ,
and
The complete solution is
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
− −= − −
− −∴ =