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**snaes** I can change this equation, using Eulers method, and get the answers using $\displaystyle sin$ and $\displaystyle cos$, but using complex numbers is apparently faster and we have to do it this way for my professor...so if you could help me figure this out I would appreciate it.

$\displaystyle y''-8y'+25y=3 e^{4x} cos(3x)$

HERE IS WHAT I DID:

First, I'll solve the homogeneous case to find the complementary solution and the roots.

$\displaystyle r^{2}-8r+25=0$

$\displaystyle r=4+3i$ or $\displaystyle r=4-3i$

$\displaystyle y_{c}=c_{1} e^{4x}cos(3x)+c_{2}e^{4x}sin(3x)$

Now, to find the solution $\displaystyle y=y_{c}+y_{p}$, where $\displaystyle y_{p}$ is the particular solution to the non-homogeneous equation, I need help. I guessed a few functions but none seemed to work out.

Here is what Ive tried:

Since $\displaystyle 3e^{4x}cos(3x)=3e^{(4+3i)x}$...well the real part I think

and

$\displaystyle 4+3i$ is a root of the homogenous equatoin, so I need multiply my guess for $\displaystyle y_{p}$ by $\displaystyle x$

From here I cannot figure out what function I need to use...