Need help finding a little equation:

Question

luigi0210 · Answer

So have the equation $\large y''+4y'+5y=35e^{-4x} $

I got the complementary solution as $\large Y_{c}=e^{-2x}(C_{1}~cosx+C_{2}~sinx) $  

Would the particular solution be $\large Y_{p}=Ae^{-4x} $ or $\large Y_{p}=(Ax+B)e^{-4x} $?

luigi0210 · Answer

The method, if it helps, is by Undetermined Coefficients - Superposition Approach

anonymous · Answer

$Y_p = Ae^{-4x}$

irishboy123 · Answer

at first glance, for the general or complimentary solution, you have complex roots and the solution is exp ( 2 +/- i) x.  right?  so i agree with that part where you extra reals using Euler.

in terms of the particular solution, you do not need to resort to Ax + B  as 4 ≠   2 +/- i.

it is only where the index of th particular solution coincides with one or more of the general solution indices that you need to start messing around.

is that any help?

luigi0210 · Answer

Yes, just wanted to make sure they didn't overlap if I used just A. Thanks for the explanation $\LARGE \color{red}{\star^{\star^{\star}}} $