TITLE: Find the particular solution of $y''+y=\cos(t)\cos(2t)$
QUESTION [5 upvotes]: Find the particular solution of $L[y]=y''+y=\cos(t)\cos(2t)$
Here my steps:
Homogeneous: $y''+y=0$
Roots $+i$ and $-i$
General solution: $y(t)=c_1\cos(t)+c_2\sin(t)+\psi(t)$
Since we are dealing with cosine then we can rewrite the right hand side as:
$y''+y=e^{it}e^{2it}=e^{{it}(1+2)}v$
Now to guess:
Let:
$\psi=e^{{it}(1+2)}v$
$\psi'=e^{{it}(1+2)}v'+3ie^{{it}(1+2)}v$
$\psi''=e^{{it}(1+2)}v''+3ie^{{it}(1+2)}v'+3ie^{{it}(1+2)}v'-9e^{{it}(1+2)}v$
Plug back into $L[y]$:
$L[y]=e^{{it}(1+2)}v''+6ie^{{it}(1+2)}v'-9e^{{it}(1+2)}v+e^{{it}(1+2)}v=e^{{it}(1+2)}v$
Dividing each side by $e^{{it}(1+2)}$
We obtain:
$L[y]=v''+6iv'-9v+v=v''+6iv'-8v=1$
Let:
$v(t)=a_0+a_1t+a_2t^2$
$v'(t)=a_1+2ta_2$
$v''(t)=2a_2$
Plug back into:
$L[y]=v''+6iv'-9v+v=v''+6iv'-8v=1$
$L[y]=2a_2+6ia_1+12ita_2-8a_0-8a_1t-8a_2t^2=1$
Equating coefficients:
$$2a_2+6ia_1-8a_0=1$$
$$ 12ia_2-8a_1=0$$
$$-8a_2=0$$
from there I got:
$$ a_2=0$$
$$ a_1=0$$
$$a_0=\frac{-1}{8}$$
Plugging back into $v(t)$
$$v(t)=\frac{-1}{8}$$
Plugging back into $\psi=e^{{it}(1+2)}v$
$$\psi=e^{{it}(1+2)}\frac{-1}{8}$$
Replacing the exponential with sine and cosines. The particular solution must be contain real values since the right hand side has cosine. 
$$\psi=[\cos(t)+i\sin(t)][\cos(2t)+i\sin(2t)]\frac{-1}{8}$$
I only multiplied the real parts:
$$\psi=\frac{-1}{8}[\cos(t)\cos(2t)-\sin(t)\sin(2t)]$$
The inside looks like the sum and angle formula so we have:
$$\psi=\frac{-1}{8}\cos(2t+t)$$
but sadly, I went wrong since the answer key says:
$\psi(t)=-\frac{1}{16}\cos(3t)+\frac{1}{4}t\sin(t)$
Honestly, I'm not sure where I went wrong

REPLY [2 votes]: We have
$\cos t \cos 2t=\frac{1}{2}(\cos3t+\cos t)$ so that $$y_h=c_1\cos t+c_2\cos t$$ and $$y_{p_1}=A\cos 3t+B\sin 3t$$and $$y_{p_2}=E\cos t+F\sin t$$ therefore $$y_p=y_{p_1}+y_{p_2}$$