Unit – 2
Linear differential equations
The general form of linear differential equation of second order is
Where p and q are constants and R is a function of x or constant.
Differential Operator
D stands for operation of differential i.e.
stands for the operator of integration.
stands for operation of integration twice.
Thus, 
Note:- Complete solution = complementary function + Particular integral
i.e. y=CF + PI
Method for finding the CF
Step1:- In finding the CF right hand side of the given equation is replaced by zero.
Step 2:- Let 
be the CF of
Putting the value of 
in equation (1) we get
It is called auxiliary equation.
Step 3:- Roots Real and Different
If 
are the roots the CF is 
If 
are the roots then 
Step 4- Roots Real and Equal
If both the roots are 
then CF is
If roots are 
Example: Solve 
Ans. Given, 
Here Auxiliary equation is
Solve: 
Or,
Ans. Auxiliary equation are 
Note: If roots are in complex form i.e. 
Solve: 
Ans. Auxiliary equation are 
Solve. 
Ans. Its auxiliary equation is
Solution is 
Solve. 
Ans. The auxiliary equation is
Hence the solution is
Rules to find Particular Integral
Case 1: 
If, 
If, 
Solve: 
Ans. Given, 
Auxiliary equation is 
Case2: 
Expand 
by the binomial theorem in ascending powers of D as far as the result of operation on 
is zero.
Solve. 
Given, 
For CF,
Auxiliary equation are 
For PI
Case 3: 
Or,
Solve: 
Ans. Auxiliary equation are 
Case 4: 
Solve. 
Ans. AE= 
Complete solution is 
Solve 
Ans. The AE is 
Complete solution y= CF + PI
Solve. 
Ans. The AE is 
Complete solution = CF + PI
Solve. 
Ans. The AE is 
Complete solutio0n is y= CF + PI
Find the PI of 
Ans. 
Solve 
Ans. Given equation in symbolic form is
Its Auxiliary equation is 
Complete solution is y= CF + PI
Solve. 
Ans. The AE is 
We know, 
Complete solution is y= CF + PI
Solve. Find the PI of (D2-4D+3)y=ex cos2x
Ans. 
Solve. (D3-7D-6) y=e2x (1+x)
Ans. The auxiliary equation i9s
Hence complete solution is y= CF + PI
A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,
The form of second order linear differential equation with constant coefficients is,
,
Where a,b,c are the constants.
Let, aD²y+bDy+cy = f(x), where d² = 
, D = 
∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy
Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)
Then we find particular integral (P.I)
P.I. = 
f(x)
General solution = C.F. +P.I.
Solve: Solve (4D² +4D -3)y = 
Auxiliary equation is 4m² +4m – 3 = 0
We get, (2m+3)(2m – 1) = 0
m = 
, 
Complementary function: CF is A
+ B
Now we will find particular integral,
P.I. = 
f(x)
= 
. 
= 
. 
= 
. 
= 
. 
= 
. 
General solution is y = CF + PI
= A
+ B
. 
Where, 
are constant is called homogenous equation.
Put, 
Solve. 
Ans. Put, 
AE is 
Solve. 
Ans. Putting, 
AE is 
CS = CF + PI
Solve. 
Ans. Let, 
AE is 
y= CF + PI
Solve. 
Ans. Let, 
so that z = log x
AE is 
Look at the following case ,
, here p and q are coefficients.
We can find the solution of these type of equation by two types of solution-
(1) general solution of the homo. Equations:
(2) particular solutions of the non- homo. Equations:
Solve: solve 
– y = 2x² - x – 3
Solution.: first we find general solution:
The characteristic function is: r² - 1 = 0
( r-1)(r+1) = 0
R = 1, -1
General solution is - A
+ B
,
Now , let
y = ax² + bx +c
= 2ax + b
2a
Put these value in 
– y = 2x² - x – 3,
2a – (ax² + bx +c) = 2x² - x – 3
2a – ax² - bx - c = 2x² - x – 3
Now compare coeff.
Coeff. Of x² , a = -2
Coeff. Of x , b = 1
Constant coeff.
2a – c = -3, c =-1
So the particular solution will be,
y = -2x² + x – 1
Complete solution is,
y = A
+ B
- 2x² + x – 1
Solve : Solve 
Here cf is r² -6r +9 = 0
(r – 3)² = 0
r = 3
So the general solution is - A
+ B
Now we will find particular solution,
Lets, y = 
= 
Substitute these values,
+ 9
= 
= 
C = 1/5
The particular solution is ,
y = 1/5 
Complete solution,
y = - A
+ B
1/5 
Working Rule
Step1: Find out the CF i.e. 
Step 2: Particular integral 
Step 3: Find u and v by formula
Solve. 
Ans. AE is 
Where,
General solution = CF + PI
Solve. (d2 y)/dx2+x2 y=sec nx
Ans. The AE is 
The Wronskian of 
is
Solve. 
Ans. Given, 
AE is 
CF is 
Complete solution is
Let us consider a resister R, an inductor L and a capacitor C which are connected in a series ,
Let ϵ(t) = ϵ’cosωt is an electromotive force to the circuit,
The equations for the voltage drops across a capacitor , a register and an inductor are
Where C is the capacitance, R is the resistance and L is the inductance , q is the charge and i is the current
We know that from Kirchhof’s law,
This is the second order non- homogeneous differential equation with constant coefficients.
Diagram for RLC circuit
Text Book:
1) Calculus: Gorakh Prasad
2) Advance Engineering Mathematics – E. Kreyszig, John Wiley & Sons Inc.
