Unit - 3
General solution of homogeneous equation of second order
Q1) What are homogeneous equation of second order differential equations?
A1)
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.
Q2) Define differential operators?
A2)
D stands for operation of differential i.e.
stands for the operator of integration.
stands for operation of integration twice.
Thus,
Q3) Solve (4D² +4D -3)y = 
A3)
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
. 
Q4) Solve 
A4)
Given, 
Here Auxiliary equation is
Q5) Solve 
A5)
Auxiliary equation is 
Q6) Solve 
A6)
Given, 
Auxiliary equation is 
Q7) 
A7)
Auxiliary equation are 
Q8) Solve 
A8)
The AE is 
Complete solution y= CF + PI
Q9) Solve 
A9)
The AE is 
Complete solution = CF + PI
Q10) Solve
A10)
The AE is 
Complete solutio0n is y= CF + PI
Q11) Solve
A11)
Given equation in symbolic form is
Its Auxiliary equation is 
Complete solution is y= CF + PI
Q12) Find the PI of (D2-4D+3)y=ex cos2x
A12)
Q13) Determine whether the following two functions are independent or dependent?
f(x) = x, g(x) = x³
A13)
Here we will make wronskian as follows,
W = 
= f(x).g’(x) – f’(x).g(x)
W = 
= 
= 3x³ - x³
Here we can see that W is not equals to zero so that these two functions are independent.
Q14) Solve 
– y = 2x² - x – 3
A14)
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
Q15) Solve 
A15)
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 
Q16) Solve the following DE by using variation of parameters-
A16)
We can write the given equation in symbolic form as-
To find CF-
It’s A.E. Is 
So that CF is- 
To find PI-
Here 
Now 
Thus PI = 
= 
= 
= 
= 
So that the complete solution is-
Q17) Find the P.I. Of (D + 2)
A17)
P.I. = 
Now we will evaluate each term separately-
And
Therefore-
Q18) Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)
A18)
Q19) Solve 
A19)
As it is a Cauchy’s homogeneous linear equation.
Put 
Then the equation becomes [D(D-1)-D+1]y = t or 
Auxiliary equation-
So that-
C.F.=
Hence the solution is- 
, we get-
Q20) Solve 
A20)
As we see that this is a Legendre’s linear equation.
Now put 
So that- 
And 
Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t
Its auxiliary equation is- 
And particular integral-
P.I. = 
Note - 
Hence the solution is - 
