(D^2+4)y=(1+Cosx)^2

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Hello, People!
Here is the video of solving higher order linear differential equation problem, whose Q is in Sinax or Cosax form.
Please have some patience and watch till end😊
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Chinnaiah Kalpana🍁
Note:
Working rule to find particular integral:
If Q=e^ax , where a is a constant.
Then,
P.I. = [ 1/f(D) ]e^ax , replace D=a (if f(a)≠0 ).
= [1/f(a)]e^ax , since f(a)≠0.
******If f(a)=0 then it is possible only when a is the root of f(m)=0.
When f(a)=0 then proceed the P.I. as follows
P.I. = x [1/f'(D)]e^ax, f'(a)≠0
= x [1/f'(a)]e^ax
Suppose that f'(a)=0 (where a is the root of twice for f(m)=0) then particuar integral becomes
P.I. = x.x [1/f''(D)]e^ax, f''(a)≠0
=x^2[1/f''(a)]e^ax
Suppose that f''(a)=0 (where a is the root of thrice for the f(m)=0) then the particular integral becomes
P.I. = x^2. x [ 1/f'''(D) ]e^ax, f'''(a)≠0
=x^3 [ 1/f'''(a) ]e^ax
and so on.
f(x)= Sinax or Cosax
If f(x)=Sinax or Cosax,
then Particular Integral is given by
P.I. = [1/Φ(D)]Sinax (or) Cosax.
In Φ(D) replace D^2 by -(a^2), provided Φ(D)≠0.
If Φ(D)=0, when we replace D^2 by -(a^2), then
P.I. = x[1/Φ'(D)]Sinax (or) Cosax.
Again replace D^2 by -(a^2) in Φ'(D) provided Φ'(D)≠0,
then
P.I.=x^2[1/Φ"(D)]Sinax (or) Cosax.
This process may be repeated if Φ"(D)=0 and so on.
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By the way mam which chapter is this?