Unit - 3
General solution of homogeneous equation of second order
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.
Let’s do an examples to understand the concept,
Differential operators-
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 
Example1: Solve (4D² +4D -3)y = 
Solution: 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
. 
Example: Solve 
Ans. Given, 
Here Auxiliary equation is
Example: Solve
Or,
Ans. Auxiliary equation are 
Note: If roots are in complex form i.e. 
Example: Solve 
Ans. Auxiliary equation is 
Rules to find Particular Integral
Case 1: 
If, 
If, 
Example: Solve 
Ans. Given, 
Auxiliary equation is 
Case 2: 
Expand 
by the
Case 3: 
Or,
Example: 
Ans. Auxiliary equation are 
Case 4: 
Example: Solve 
Ans. AE= 
Complete solution is 
Example: Solve 
Ans. The AE is 
Complete solution y= CF + PI
Example: Solve 
Ans. The AE is 
Complete solution = CF + PI
Example: Solve
Ans. The AE is 
Complete solutio0n is y= CF + PI
Example: Find the PI of 
Ans. 
Example: solve
Ans. Given equation in symbolic form is
Its Auxiliary equation is 
Complete solution is y= CF + PI
Example: Solve
Ans. The AE is 
We know, 
Complete solution is y= CF + PI
Example: Find the PI of(D2-4D+3)y=ex cos2x
Ans. 
Example. Solve(D3-7D-6) y=e2x (1+x)
Ans. The auxiliary equation i9s
Hence complete solution is y= CF + PI
Key takeaways-
- The second order linear differential equation with constant coefficients is,
Where a,b,c are the constants.
2. General solution = C.F. +P.I.
3. 
4. 
5. Roots Real and Equal-
6. Roots Real and Different-
7. If roots are in complex form i.e. 
If we have two functions f(x) and g(x) then wronskian is,
W = 
= f(x).g’(x) – f’(x).g(x)
If the wronskian is equal to zero then the two functions are dependent , if wronskian is not equal to zero the two functions are independent.
Example: Determine whether the following two functions are independent or dependent?
f(x) = x, g(x) = x³
Sol. 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.
Example: Determine whether the following two functions are independent or dependent?
f(x) = sinx, g(x) = cosx
Sol.
Here we will make wronskian as follows,
W = 
= f(x).g’(x) – f’(x).g(x)
W = 
= -sinx * sinx – cosx*cosx
= - sin²x - cos²x
= -( sin²x + cos²x) = -1
Which is not equal to zero, hence we can say that the given function is not dependent.
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:
Example: 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
Example: 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 
Method of variation of parameters-
Consider a second order LDE with constant co-efficients given by
Then let the complimentary function 
is given by
Then the particular integral is
Where u and v are unknown and to be calculated using the formula
u=
Example-1: Solve the following DE by using variation of parameters-
Sol. 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-
Example-2: Solve the following by using the method of variation of parameters.
Sol. This can be written as-
C.F
Auxiliary equation is- 
So that the C.F. Will be- 
P.I.-
Here 
Now 
Thus PI = 
= 
= 
So that the complete solution is-
Linear differential equations are those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.
Thus the general linear differential equation of the n’th order is of the form
Where 
and X are function of x.
Linear differential equation with constant co-efficient are of the form-
Where 
are constants.
Rules to find the complementary function-
To solve the equation-
This can be written as in symbolic form-
Or-
It is called the auxiliary equation.
Let 
be the roots-
Case-1: If all the roots are real and distinct, then equation (2) becomes,
Now this equation will be satisfied by the solution of 
This is a Leibnitz’s linear and I.F. = 
Its solution is-
The complete solution will be-
Case-2: If two roots are equal 
Then complete solution is given by-
Case-3: If one pair of roots be imaginary, i.e. 
then the complete solution is-
Where 
and 
Case-4: If two points of imaginary roots be equal-
Then the complete solution is-
Example-Solve 
Sol.
Its auxiliary equation is-
Where-
Therefore the complete solution is-
Inverse operator-
is that function of x, not containing arbitrary constants which when operated upon gives X.
So that 
satisfies the equation f(D)y = X and is, therefore, its P.I.
f(D) and 1/f(D) are inverse operator.
Note-
1. 
2. 
Rules for finding the particular integral-
Let us consider the equation-
Or in symbolic form-
So that-
Now-
Case-1: When X = 
In case f(a) = 0, then we see that the above rule will not work,
So that-
Example: Find the P.I. Of (D + 2)
Sol.
P.I. = 
Now we will evaluate each term separately-
And
Therefore-
Example: Solve (D – D’ – 2 ) (D – D’ – 3) z = 
Sol.
The C.F. Will be given by-
Particular integral-
Therefore the complete solution is-
Case-2: when X = sin(ax + b) or cos (ax + b)
In case 
then the above rule fails.
Now-
And if 
Similarly-
Example: Find the P.I. Of 
Sol.
Example: Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)
Sol.
Example: Find P.I. Of
Sol. P.I = 
Replace D by D+1
Put 
Key takeaways-
- If all the roots are real and distinct-
The complete solution will be-
2. If two roots are equal 
Then complete solution is given by-
3. If one pair of roots be imaginary, i.e. 
then the complete solution is-
Where 
and 
4. If two points of imaginary roots be equal-
An equation of the form
Here X is the function of x, is called Cauchy’s homogeneous linear equation.
Example-1: Solve 
Sol. 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-
Example-2: Solve 
Sol. On putting 
so that,
and
The given equation becomes-
Or it can be written as-
So that the auxiliary equation is- 
C.F. = 
Particular integral- 
Where 
It’s a Leibnitz’s linear equation having I.F.=
Its solution will be-
P.I. = 
= 
So that the complete solution is-
An equation of the form-
Is called Legendre’s linear equation.
Example-3: Solve 
Sol. 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 - 
References:
- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
