Question ODE non-homogeneous Linear

[Another]

Mod note: Member warned that the homework template is NOT optional
find y_p (particular integral)

(D² + 4D + 5) y = 2 e^-2xcos(x)

((D+2)(D+2)+1) y = 2 e^-2xcos(x)

y_p = [1/((D+2)² + 1)] ⋅ 2e^-2xcos(x)

y_p = e^(-3)x ∫ ∫ e^-(-3)x⋅2e^-2xcos(x)dxdx

y_p = e^(-3)x ∫ ∫ e^3x ⋅ 2e^-2xcos(x)dxdx

y_p = e^(-3)x ∫ (e^xsin(x)+e^xcos(x))dx

y_p = e^(-3)x[-½ e^xcos(x) + ½e^xsin(x) + ½e^xsin(x) + ½e^xcos(x)]

y_p = e^(-3)x ⋅ e^x sin(x)

y_p = e^(-2)x sin(x) ; this is my answer

But correct answer is x e^(-2)x sin(x)

Why are there variables x ?

 

[Orodruin]

Did you try inserting your solution into the ODE to check if it solves it or not?

Did you try to insert the given solution into the ODE to check if it solves it or not?

 

[Ray Vickson]

find y_p (particular integral)

(D² + 4D + 5) y = 2 e^-2xcos(x)

((D+2)(D+2)+1) y = 2 e^-2xcos(x)

y_p = [1/((D+2)² + 1)] ⋅ 2e^-2xcos(x)

y_p = e^(-3)x ∫ ∫ e^-(-3)x⋅2e^-2xcos(x)dxdx

y_p = e^(-3)x ∫ ∫ e^3x ⋅ 2e^-2xcos(x)dxdx

y_p = e^(-3)x ∫ (e^xsin(x)+e^xcos(x))dx

y_p = e^(-3)x[-½ e^xcos(x) + ½e^xsin(x) + ½e^xsin(x) + ½e^xcos(x)]

y_p = e^(-3)x ⋅ e^x sin(x)

y_p = e^(-2)x sin(x) ; this is my answer

But correct answer is x e^(-2)x sin(x)

Why are there variables x ?

You have not found a particular solution; you have found one of the two homogeneous solutions (that is, solutions which solve the DE with 0 on the right).

In general, I am extremely suspicious of your approach, because I do not trust expressions like ##[(D+2)^2+1]^{-1} f(x)##. You seem to be attempting to use Heaviside's Operational Calculus, but without sticking to the rules.

It would be better to use either
(1) Laplace transforms --- basically, a modern version of Heavisides's operational methods; or
(2) The method of undetermined coefficients.
Personally, I prefer (1), and it leads to exactly the particular solution ##x e^{-2x} \sin x## that somebody has told you is correct.

I am not sure how to answer your question about "why the x?". All I can say is that the two homogenous solutions ##y_1(x) = e^{-2x} \cos x## and ##y_2(x) = e^{-2x} \sin x## do not work, so trying something like ##c_1(x) y_1(x) + c_2(x) y_2(x)## is the obvious next step. If you look to make ##c_1(x)## and ##c_2(x)## as simple as possible, you would try linear functions ##c_i(x) = a_i + b_i x##, and then try to determine the ##a_i## and ##b_i##. That is more-or-less method (2) that I mentioned above. See, eg., https://en.wikipedia.org/wiki/Method_of_undetermined_coefficients .