1) The document discusses second order linear homogeneous differential equations, which have the general form P(x)y'' + Q(x)y' + R(x)y = 0.
2) It describes methods for finding the general solution including reduction of order, and discusses the solutions when the coefficients are constants.
3) The general solution depends on the nature of the roots of the auxiliary equation: distinct real roots, repeated real roots, or complex roots.
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Second order homogeneous linear differential equations
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
駕
7. let c = 1,
: independent solutions
7
( )
0
G x dx
u v e
駕
( )
( )
G x dx
u x dxe
駕
( )
2 1 1( ) ( ) ( ) ( )
G x dx
y x u x y x y x dxe
駕
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
緒
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
緒
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
緒
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=piq
e(p+iq)x, e(piq)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