Given the differential equation:

$\displaystyle \frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=u '+3u$

Find the $\displaystyle y(t)$ solution when $\displaystyle u(t)=e^{t}\cos t$

Hint: $\displaystyle e^{t}\cos t=Re(e^{(1+i)t})$

The answer should be:

$\displaystyle y(t)=\frac{32}{195}e^{t}\cos t+\frac{9}{195}e^{t}\sin t$

My attempt (I'm not sure if I'm using the correct method):

I start by using the "guessing method" which states that if you have $\displaystyle u(t)=sin(kt)$ then you can guess on the solution with the form:

$\displaystyle y(t)=A \cos(kt)+B \sin(kt)$

So my guess so far is:

$\displaystyle y(t)=A \cos(t)+B \sin(t)$

I hereby differentiate this guess 2 and 4 times and insert it to my original equation:

$\displaystyle y''(t)=-A \cos(t)-B \sin(t)$

$\displaystyle y''''(t)=A \cos(t)+B \sin(t)$

Inserted:

$\displaystyle (A \cos(t)+B \sin(t))+6\cdot (-A \cos(t)-B \sin(t))+25\cdot (A \cos(t)+B \sin(t))=u'+3u$

I clean it up a little:

$\displaystyle 20\cdot A \cos(t)+20\cdot B \sin(t)=u'+3u$

This is as far as I've made it because in the examples in my textbook, they have not shown what to do when $\displaystyle u$ appears in the original equation, thus I have no idea how to isolate and find the values of $\displaystyle A$ and $\displaystyle B$.

Can somebody please tell me if I'm on right track at all, and if so, what do I do next?