This document discusses homogeneous and non-homogeneous differential equations. It defines homogeneous linear second-order differential equations and explains their standard form. It also discusses methods to solve homogeneous equations with real or complex roots, including repeated roots. For non-homogeneous equations, it explains that the general solution has two parts: the complementary function (solution of the homogeneous part) and a particular solution. It provides examples of guessing particular solutions for different non-homogeneous terms like constants, polynomials, and trigonometric functions.

2. HOMOGENEOUS
DIFFERENTIAL EQUATION
Standard form of homogeneous differential
equation :
y’’+ay’+by=0; a,b are real numbers

3. A differential equation of the type
ay’’+by’+cy=0, a,b,c real numbers,
is a homogeneous linear second order differential equation.
SECOND ORDER DIFFERENTIAL
EQUATIONS
Definition
     2
e and e .mx mx mx
y y me y m
           2 2
0 e e e 0 0.mx mx mx
ay by cy am bm c am bm c
To solve the equation ay’’+by’+cy=0 substitute y = emx and try to determine m so that
this substitution is a solution to the differential equation.
Compute as follows:
Homogeneous linear second order differential equations can always be solved by
certain substitutions.
This follows since emx≠0 for all x.
Definition The equation am2 + bm+ c = 0 is the Characteristic Equation of the
differential equation ay’’ + by’ + cy = 0.

4. SOLVING HOMOGENEOUS 2ND ORDER
LINEAR EQUATIONS: CASE I
Equation
Case I
 1 2
1 2e em x m x
y C C
CE has two different real solutions m1 and m2.
ay’’+by’+cy=0 CE am2+bm+c=0
In this case the functions y = em1x and y = em2x are both solutions to the original
equation.
General Solution
Exampl
e
0y y   CE  2
1 0m    1 or m 1.m
1 2e ex x
y C C 
 General Solution
The fact that all these functions are solutions can be verified by a direct calculation.

5. SOLVING HOMOGENEOUS 2ND ORDER
LINEAR EQUATIONS: CASE II
Equation
Case II
 1 2e emx mx
y C C x
CE has real double root m.
ay’’+by’+cy=0 CE
In this case the functions y = emx and y = xemx are both solutions to the original
equation.
General Solution
Example 2 0y y y    CE   2
2 1 0m m   1 (double root).m
 1 2e ex x
y C C xGeneral Solution
am2+bm+c=0

6. SOLVING HOMOGENEOUS 2ND ORDER
LINEAR EQUATIONS: CASE III
Equation
Case III
    
  1 2e sin e cosx x
y C x C x
ay’’+by’+cy=0 CE
General Solution
Example 2 5 0y y y   
CE   2
2 5 0m m    1 2m i
   
   1 2e sin 2 cos 2x
y C x C xGeneral Solution
CE has two complex solutions    .m i
In this case the functions
are both solutions to the original equation.
    
  e sin and e cosx x
y x y x
am2+bm+c=0
    
  1 2e sin cosx
y C x C x

7. REAL AND UNEQUAL ROOTS
 If roots of characteristic polynomial P(m) are real and
unequal, then there are n distinct solutions of the
differential equation:
 If these functions are linearly independent, then
general solution of differential equation is
 The Wronskian can be used to determine linear
independence of solutions.
21
, , , nm x m xm x
e e eK
21
1 2( ) nm x m xm x
ny x c e c e c e   K

8. EXAMPLE 1: DISTINCT REAL ROOTS (1 OF 3)
 Solve the differential equation
 Assuming exponential soln leads to characteristic
equation:
 Thus the general solution is
    
4 3 2
( ) 2 13 14 24 0
1 2 3 4 0
mx
y x e m m m m
m m m m
      
     
2 3 4
1 2 3 3( ) x x x x
y x c e c e c e c e 
   
(4)
2 13 14 24 0y y y y y      

9. COMPLEX ROOTS
 If the characteristic polynomial P(r) has complex roots,
then they must occur in conjugate pairs,
 Note that not all the roots need be complex.
i 
In this case the functions
are both solutions to the original equation.
    
  e sin and e cosx x
y x y x
    
  1 2e sin cosx
y C x C x
General Solution

10. EXAMPLE 2: COMPLEX ROOTS
 Consider the equation
 Then
 Now
 Thus the general solution is
  3 2
( ) 1 0 1 1 0mt
y t e m m m m        
0 yy
2 1 1 4 1 3 1 3
1 0
2 2 2 2
i
m m m i
    
        
   /2
1 2 3( ) cos 3 / 2 sin 3 / 2x x
y t c e e c x c x    
 

11. EXAMPLE 3: COMPLEX ROOTS (1 OF 2)
 Consider the initial value problem
 Then
 The roots are 1, -1, i, -i. Thus the general solution is
 Using the initial conditions, we obtain
 The graph of solution is given on right.
  4 2 2
( ) 1 0 1 1 0mx
y x e r r r       
2)0(,2/5)0(,4)0(,2/7)0(,0)4(
 yyyyyy
   1 2 3 4( ) cos sinx x
y x c e c e c x c x
   
   
1
( ) 0 3 cos sin
2
x x
y x e e x x
   

12. REPEATED ROOTS
 Suppose a root m of characteristic polynomial P(r) is a
repeated root with multiplicity n. Then linearly
independent solutions corresponding to this repeated
root have the form
2 1
, , , ,m x m x m x s mx
e xe x e x e
K

13. EXAMPLE 4: REPEATED ROOTS
 Consider the equation
 Then
 The roots are 2i, 2i, -2i, -2i. Thus the general solution
is
  4 2 2
( ) 8 16 0 4 4 0mx
y x e m m m m        
0168)4(
 yyy
   1 2 3 4( ) cos2 sin2 cos 2 sin 2y x c x c x c x x c x x   

15. 15
Two solutions are linearly independent.
Their linear combination provides an
infinity of new solutions
○ Definition 2.1: f , g : linearly dependent
If s.t. or ;
otherwise f , g : linearly independent
In other words, f and g are linearly
dependent only if for
,c ( ) ( )f x cg x ( ) ( )g x cf x
※
1 2 0c f c g 1 2 0c c 

16. ○ 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
16
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

 

17. ○ Let
If : linear dep., then or
Assume
17
1 1 2 2 0 ( )c y c y A      
1 1 2 2 0 ( )c y c y B    
1 1 2 1 2 1 2
(A) ( ) ( ) 0y B y c y y y y       
2 2 1 1 2 1 2
(A) ( ) ( ) 0,y B y c y y y y       
1 2,y y
1 0c  2 0c 
1 1 2 1 20 0c y y y y    

18. 。 Example 2.3:
Solve by a power series method
The Wronskian of at nonzero x
would be difficult to evaluate, but at x = 0
are linearly independent
18
0y xy 
3 6 9
1
4 7 10
2
1 1 1
( ) 1
6 180 12960
1 1 1
( )
12 504 45360
y x x x x
y x x x x x
     
     
1 2,y y
1 2
1 2
(0) (0) 1 0
(0) 1 0
(0) (0) 0 1
y y
W
y y
   
 
1 2,y y

19.  The general solution of the non homogeneous
differential equation
There are two parts of the solution:
1. solution of the homogeneous part of DE
2. particular solution
( )ay by cy f x   
cy
py

20.  General solution
c py y y 
Complementary Function,
solution of Homgeneous part
Particular Solution

21. The method can be applied for the non – homogeneous differential equations ,
if the f(x) is of the form:
1. A constant C
2. A polynomial function
3.
4.
5. A finite sum, product of two or more functions of type (1- 4)
( )ay by cy f x   
mx
e
sin ,cos , sin , cos ,...x x
x x e x e x 
   

22. ○ Guess from
Let : a given polynomial
, : polynomials with unknown coefficients
22
py ( )R x
( )P x
( )Q x ( )S x
Guessed( )R x py
( )P x ( )Q x
ax
ce ax
de
cos sinbx or bx  cos sinc bx d bx
( ) ax
P x e ( ) ax
Q x e
( )cos ( )sinP x bx or P x bx ( )cos ( )sinQ x bx S x bx
( ) cos ( ) sinax ax
P x e bx or P x e bx ( ) cos ( ) sinax ax
Q x e bx S x e bx

23. APPLICATIONS

27. PREPARED BY;
NAME ENROLLMENT NO.
PARMAR KALPESH 160280102031
PATEL DIPANG N. 160280102033
PATEL DRUMIL R. 160280102034
PATEL HARSH P. 160280102035
PATEL JAY 160280102036
PATEL KARMIT R. 160280102037
PATEL NIKUNJ 160280102039