My DE equation looks like this:
$y^{(3)}+y^{(2)}+34y^{(1)}+40y=xe^{-4x}+2e^{-3x}cos(x)$
I'm having trouble solving for the characteristic equation
$r^4+r^3+34r^2+40r=0$
I got it down to $r(r^3+r^2+34r+40)=0$ but I'm not sure how to factor it any further to find real zeros. I guess this is more of an algebra type ordeal, but I'm stumped.
I bet that your equation is not $y^{(3)}+y^{(2)}+34y^{(1)}+40y=xe^{-4x}+2e^{-3x}cos(x)$, but is : $$y^{(3)}+10y^{(2)}+34y^{(1)}+40y=xe^{-4x}+2e^{-3x}cos(x)$$ Hit : $r^3+10r^2+34r+40=(r+4)(r+3+i)(r+3-i)$
Do you understand the reason of the guess ?