2. Linear Differential Equations of Second Order
• The general second order Linear Differential Equation is
or
where P(x) ,Q(x) and R (x) are functions of
only.
2
2
( ) ( ) ( )
d y dy
P x Q x y R x
dx dx
( ) ( ) ( ) (1)y P x y Q x y R x
x
3. • The term R(x) in the above equation is isolated from
others and written on right side because it does not
contain the dependent variable y or any of its derivatives.
•If R(x) is Zero then,
•The solution of eq.(2) which is homogeneous linear
differential equation is given by,
( ) ( ) 0 (2)y P x y Q x y
1 1 2 2y c y c y
4. where c1 and c2 are arbitary constants
• Two solutions are linearly independent .Their linear
combination provides an infinity of new solutions.
•Wronskian test - Test whether two solutions of a
homogeneous differential equation are linearly independent.
Define: Wronskian of solutions to be the 2
by 2 determinant
1 2,y y
1 2 1 2( ) ( ) ( ) ( ) ( )W x y x y x y x y x
1 2
1 2
( ) ( )
( ) ( )
y x y x
y x y x
5. • Example:- are solutions of
:linearly independent.
Reduction of Order
A method for finding the second independent
homogeneous solution when given the first one .
1 2( ) cos , ( ) siny x x y x x
0y y
1 2
1 2
2 2
( ) ( ) cos sin
( )
( ) ( ) sin cos
cos sin 1 0
y x y x x x
W x
y x y x x x
x x
1 2,y y
2y 1y
6. • Let
Substituting into
Let (separable)
6
2 1( ) ( ) ( )y x u x y x
2 1 1 2 1 1 1, 2y u y uy y u y u y uy
( ) ( ) 0y P x y Q x y
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1
2 [ ] 0
[2 ] [ ] 0
[2 ] 0
u y uy P u y uy Quy
u y Py u y Py Qy
u y Py
u y
u y
u y
1 1
1
2
0 ( ) 0
y Py
u u u G x u
y
( ) 0v u v G x v
( )
( )
G x dx
v x ce
7. let c = 1,
: independent solutions
7
( )
0
G x dx
u v e
( )
( )
G x dx
u x dxe
( )
2 1 1( ) ( ) ( ) ( )
G x dx
y x u x y x y x dxe
1 2 1 2 1 1 1 1 1
2 2
1 11 1 1 1 1 1
( )
0
( ) y y y y uy u y y uy
u y vy
W x y
y uy y u y y uy
1 2 1,y y uy
。 Example 2.4:
: a solution
Let
4 4 0, (3)y y y
2
( ) x
y x e
2
2 1( ) ( ) ( ) ( ) x
y x u x y x u x e
8. Substituting into (3),
take c = 1, d = 0
: independent
The general solution:
8
2 2
2 2 ,x x
y u e e u
2 2 2
2 4 4x x x
y u e e u u e
2 2 2 2 2 2
4( ) 4 04 4 2x x x x x x
uu e e u u e u e e u e
2 2
0, 0, 0, ( )x x
u x cx du e e u
( )u x x
2 2
2 ( ) ( ) x x
y x u x e xe
2 2
4
2 2 2
0
2
( )
2
x x
x
x x x
x
e xe
W x e
e e e
2 2
1 2( ) x x
c xy x c e e
2 2
1 2,x x
y y xe e
9. Second order Linear Homogeneous Differential
Equations with constant coefficients
a,b are numbers ------(4)
Let
Substituting into (4)
( Auxilliary Equation)
--------(5)
The general solution of homogeneous D.E. (4) is
obtained depending on the nature of the two roots of the
auxilliary equation as follows :
0y ay by
( ) mx
y x e
2
0mx mx mx
m am be e e
2
0m am b
2
1 2, 4 ) / 2(m a a bm
10. Case 1 : Two distinct real roots if and only if
in this case we get two solutions &
Solution is
Case 2 : Equal real roots if
Solution is
Case 3 : Distinct complex roots if and only if
In this case & can be written in the form of
Solution is
2
4 0a b
1m x
e 2m x
e
1 2
1 2( ) m x m x
y x c ce e
2
4 0a b
1 2 1 2( ) ( )mx mx mx
y x c c x c c xe e e
2
4 0a b
1m 2m
p iq
1
2( ) [ cos sin ]px
y x e c qx c qx
11. Case Roots Basis of solutions General Solution
1 Distinct real m1 and
m2
em1x, em2x c1em1x+c2em2x
2 Repeated root m emx, xemx c1emx+c2xemx
3 Complex roots
m1=p+iq, m2=p–iq
e(p+iq)x, e(p–iq)x epx(c1cos qx+c2sin qx)
Example for Case 1 :
Let then,
From (6) ,
The general solution:
6 0 (6)y y y
( ) mx
y x e 2
,mx mx
y me y m e
2 2
6 0 6 0mx mx mx
m m m me e e
1 2( 2)( 3) 0 2, 3m m m m
2 3
1 2( ) x x
y x c ce e
12. Example for Case 2 :
Characteristic eq. :
The repeated root:
The general solution:
Example for Case 3 :
Characteristic equation:
Roots:
The general solution:
6 9 0y y y
2 2
6 9 0 ( 3) 0m m m
3m
3
1 2( ) ( ) x
y x c c x e
2 6 0y y y
2
2 6 0m m
1 21 5 , 1 5m i m i
( 1 5 ) ( 1 5 )
1 2( ) i x i x
y x c ce e
