## cke8 2 years ago Please help me solve y''(x)+6 y'(x)+13 y(x) = 4 e^(-3 x) sin(2 x), y(0) = 1, y'(0) = 2

1. SithsAndGiggles

$y''+6y'+13y=4e^{-3x}\sin(2x),\; y(0)=1,\; y'(0)=2$ Homogeneous solution: $r^2+6r+13=0\\ r=\frac{-6\pm\sqrt{-16}}{2}=-3\pm4i\\ y_c=e^{-3x}\left(C_1\cos(4x)+C_2\sin(4x)\right)$ Non-homogeneous solution: Using the method of undetermined coefficients, try (as a guess) $y_p=Ae^{-3x}\sin(2x)+Be^{-3x}\cos(2x)$ I'll leave that work to you, since it'd be a pain to type it all out, but I'll be available to check what you get later. The general solution will then be $y=y_c+y_p$

2. cke8

Could you please explain a bit more?

3. abb0t

I think the explanation provided is quite clear.

4. SithsAndGiggles

@cke8, which part? The homogeneous solution, or the clue for the non-homogeneous one?

5. cke8

The clue for the non-homogenous one. Thanks!

6. SithsAndGiggles

\begin{align*}y_p&=Ae^{-3x}\sin(2x)+Be^{-3x}\cos(2x)\\ &=e^{-3x}\left(A\sin(2x)+B\cos(2x)\right)\\\\ y_p'&=-3e^{-3x}\left(A\sin(2x)+B\cos(2x)\right)+e^{-3x}\left(2A\cos(2x)-2B\sin(2x)\right)\\ &=e^{-3x}\left[(-3A-2B)\sin(2x)+(-3B+2A)\cos(2x)\right]\\\\ y_p''&=-3e^{-3x}\left[(-3A-2B)\sin(2x)+(-3B+2A)\cos(2x)\right]\\&\;\;\;\;\;+e^{-3x}\left[(-6A-4B)\cos(2x)+(6B-4A)\sin(2x)\right]\\ &=e^{-3x}[(5A+12B)\sin(2x)+(5B-12A)\cos(2x)] \end{align*} Now plug in y_p'', y_p', and y_p into the original equation and solve for A and B. It seems daunting, but doable. Be sure to keep track of your coefficients and their signs.