Second order non homogeneous diff. equation at constant coefficients

[Telemachus]

Hi there. I had some trouble trying to solve this:

y''+y=\cos x +3\sin 2x (1)

At first I just found the solution for the homogeneous equation:
y_h=e^{\lambda x} \rightarrow \lambda^2+1=0 \rightarrow \lambda_1,\lambda_2=\pm i

Then y_h=C_1\cos x+ C_2 \sin x

So I've tried to find the particular solution. I thought I should suggest an equation like:
y_p=A\cos x +b\sin x+ C\cos 2x +D\sin 2x
y_p'=-A\sin x +B\cos x - 2C\sin 2x + 2D\cos 2x
y_p''=-A\cos x -B\sin x - 4C\cos 2x - 4D\sin 2x
But then, when I've tried to find the undetermined coefficients for y_p coefficients replacing in (1):

-A\cos x -B\sin x - 4C\cos 2x - 4D\sin 2x+A\cos x +b\sin x+ C\cos 2x +D\sin 2x=\cos x +3\sin x \rightarrow -3C\cos 2x -3D\sin 2x=cos(x)+3\sin 2x
Then A=0,B=0,C=0 and D=-1.

So the general solution should be
y(x)=y_h+y_p=C_1\cos x+ C_2 \sin x-\sin 2x
But with wolfram alpha I've corroborated my solution is wrong: y''+y=cos(x)+3 sin(2x) - Wolfram|Alpha

So, where is the mistake and how should I do this?

I think I've found unless one mistake, just noted it. I should use y_p=Ax \cos x +B x \sin x + C\sin 2x+ D\cos 2x instead of y_p=A\cos x +b\sin x+ C\cos 2x +D\sin 2x, because the first two terms are linearly dependent with the homogeneous solution, right?

 

[Mark44]

For your particular solution, try
y_p=Ax\cos x +Bx\sin x

The problem here is that cos(x) and sin(x) are solutions of the homogeneous version of your differential equation, so a particular solution (of the nonhomogeneous equation) can't include either of these functions.

 

[Telemachus]

right, I've made a mistake in (1) that I've just corrected, it was 2x for the sine instead of x. Shouldn't I add some more terms for the particular solution corresponding to the part of the equality of 3\sin 2x?

 

[Mark44]

For this nonhomogeneous equation,
y''+y=\cos x +3\sin 2x

try this for your particular solution,
y_p=Ax\cos x +Bx\sin x + C\cos 2x + D\sin 2x

 

[Telemachus]

Thank you Mark :)

 

[Mark44]

De nada...

I didn't go into any detail about why I chose the functions I did. For relatively simple nonhomogenous, constant coefficient, linear differential equations, the concept of annihilators provides some insight into the choices for the particular solutions.

In this thread, https://www.physicsforums.com/showthread.php?t=348247, in post #4, I talk about it a bit.

 

[Telemachus]

Muchas gracias :D