Find the P.I of (D+1) ^2 = x
Answers
Answer:
If both m1 and m2 are constants, the expressions (D – m1) (D – m2) y and (D– m2) (D – m1) y are equivalent i.e. the expression is independent of the order of operational factors.

We will explain the method with the help of following
Example: Solve d2y/ dx2 – 5 dy/dx +6y = e3x.
Solution:
The equation can be written as
(D2 – 5D + 6)y = e3x
(D – 3) (D – 2)y = e3x
C.F. = c1 e3x + c2e2x
And P.I. =1/ (D-3). 1/ (D-2) e3x
= 1/ (D-3) e2x ∫ e3x e-2xdx
=1/ (D-3) e2x ex
= e3x ∫ e3x e-3x dx = x.e3x
y = c1e3x +c2e2x +xe3x
P.I. can be found by resolving
1/f (D) = 1/ (D-3). 1/ (D-2)
Now using partial fractions,
1/f (D) = 1/ (D-3). 1/ (D-2)
= 1/ (D-3) – 1/ (D-2)
Hence, the required P.I. is [1(D-3) – 1/ (D-2)] e3x
= 1/ (D-3) e3x – 1/ (D-2) e3x
= e3x ∫ e3x e-3x dx – e2x ∫ e3x e-2x dx
= xe3x – e3x
Second term can be neglected as it is included in the first term of the C.F.
Short Method of Finding P.I.
In certain cases, the P.I. can be obtained by methods shorter than the general method.
(i). To find P.I. when X = eax in f(D) y = X, where a is constant
y = 1/f(D)
1/f(D) eax = 1/f(a) eax , if f(a) ≠ 0.
1/f(D) eax = xr/f r(a) eax , if f(a) = 0, where f(D) = (D-a)rf(D)
Example: Solve (D3 – 5D2 + 7D – 3)y = e3x.
Solution:
(D – 1)2 (D – 3) y = e3x
C.F. = aex + bx ex + ce3x
And P.I. = 1/ (D3-5D2+7D-3) e3x
= x/ (3D2-10D+7) e3x
= ¼ xe3x
y = aex + bx ex + ce3x + ¼ xe3x.
(ii). To find P.I. when X = cos ax or sin ax
f (D) y = X
y = 1/f(D) sinax
If f (– a2) ≠ 0 then 1/f(D2) sinax = 1/f(-a2) sinax
If f (– a2) = 0 then (D2 + a2) is at least one factor of f (D2)
Let f (D2) = (D2 + a2)r f (D2)
Where f (– a2) ≠ 0
[1/f (D2)] sin ax = 1/ (D2+a2). 1/φ(D2) sin ax
= 1/ φ(-a2). 1/ (D2+a2) sin ax
Now, when r=1, 1/(D2 + a2) sin ax = -x/2a. cos ax
Similarly If f(–a)2 ≠ 0 then 1/f(D2) cos ax = 1/f(-a2) cosax
And 1/ (D2+a2) cos ax = x/2a .sin ax
Example: Solve (D2 – 5D + 6) y = sin3x.
Solution: We can factorize the given term as (D – 2) (D – 3)y = sin3x
C.F. = ae2x + be3x
P.I. = 1/ (D2 -5D +6) sin 3x
= 1/ [(-9) -5D +6] sin 3x
= 1/ (-5D-3) sin 3x
= - (5D-3). 1/ (25D2 -9) sin3x
= 1/234 (15 cos 3x-3 sin3x)
= ae2x + be3x + 1/234 (15 cos 3x-3 sin3x)
(iii) To find the P.I. when X = xm where m ∈ N
f (D) y = xm
y = 1/ f(D) xm
We will explain the method by taking an example
Example: Find P.I. of (D3 + 3D2 + 2D) y = x2.
Solution: The P.I. is given by 1/ (D3+3D2+2D) x2
=  = 
= 
= 1/2D (1-3D/2+7/4D2 + ….) x2
= 1/2D (x2-3x +7/2)
And 1/2D (x2-3x+7/2) = 1/2(x3/3-3x2/2+7/2x) = 1/12 (2x3 -9x2+21x).
(iv) To find the value of 1/f(D) eax V where ‘a’ is a constant and V is a function of x
1/f (D) .eax V = eax.1/f (D+a). V
Example: Solve (D2 + 2) y = x2 e3x.
Solution: C.F. = a cos √2x + b sin √2x
P.I. = 1/ (D2+2). x2e3x = e3x. 1/ [(D+3)2 +2] .x2
= e3x. 1/(D2 +6D+11). x2
= 1/11. e3x (1+ (6D+D2)/11)-1 x2
= 1/11 e3x [1- (6D+D2)/11 + 36/121 . D2 +…) x2
= 1/11 e3x (1-6D/11+25/121 . D2+…) x2
= 1/11 e3x (x2 – 12/11 x + 50/121)
= a cos √2x + b sin √2x + e3x/11 (x2-12/11 x +50/121)
(v). To find 1/f (D). xV where V is a function of x
1/f (D). xV = [x- 1/f(D). f'(D)] 1/f(D) V