Solving $y''+2y'+2y=e^{-x}\sin(x)$ using variation of parameters.

Question

I am stuck on the following differential equation. Here is where I got to. We first solve the homogeneous equation. \begin{align*} &y^{\prime\prime} + 2y^{\prime} + 2y = 0\Longrightarrow r^2+2r+2=0 \Longrightarrow \sqrt{D} = \sqrt{2^2-4\cdot1\cdot2} = \pm2i \Longrightarrow\\ &r = \frac{-2\pm2i}{2}=-1\pm i \Longrightarrow y = Ae^{-x}\cos(x)+Be^{-x}\sin(x) \end{align*} Now I vary the parameter: $y(x)=A(x)e^{-x}\cos(x)+B(x)e^{-x}\sin(x)$. But from here I have no idea how to plug the non-homogeneous equation. Do I need to use a 'wronskian' as noted in my book or simply take derivatives?

Answer 1

If you want to use the Undetermined Coefficients, you should take few steps ahead just to find the proper particular solution $y_p$. It seems to me that you are walking on this railway. But, if you are interested in doing the Variation of Parameters so use the Wronskian as guessed. So, find the wronskian of $$y_1=e^{-x}\sin(x), ~y_2=e^{-x}\cos(x)$$ which is $W=-e^{-2x}(\neq 0)$ and for the rest use the formal relevant formulas: $$y_p=u_1y_1+u_2y_2$$ where $$u_1=\int\frac{-y_2\times e^{-x}\sin(x)}{W}dx, ~~u_2=\int\frac{y_1\times e^{-x}\sin(x)}{W}dx$$

Answer 2

The Wronskian is the way to go.

Otherwise you have to deal with a big mess of calculations.

Make sure that you check your answer and make sure you had the right signs in your formula for your coefficients.

Answer 3

Let $u = Ae^{-x}\cos x + Be^{-x}\sin x$

We know that $u'' + 2u' + 2u = 0$

Suppose $y = xu$

Then

$y' = u + xu'\\ y'' = 2u' + xu''$

$y'' + 2y' + 2y = 2u' + 2u + x(u'' + 2u' + 2u) = e^{-x}\sin x\\ 2(u'+u) = e^{-x}\sin x\\ u' = -Ae^{-x}\cos x - Ae^{-x}\sin x - Be^{-x}\sin x + Be^{-x}\cos x\\ 2(u'+u) = - 2Ae^{-x}\sin x +2Be^{-x}\cos x = e^{-x}\sin x \\ B = 0, A = -\frac 12 $

$y = C_1e^{-x}\cos x + C_2e^{-x}\sin x - \frac 12 e^{-x}\cos x$