Given the differential equation $y'' +2y'+5y = 3e^{-x}\sin(x)$, with $y(0) = 0$, $y'(0) = 3$, I'm asked to use Laplace transforms to find the solution.
So I used the tables to get the following:
$\begin{align}\mathcal{L}[y]''&=s^2\mathcal{L}[y] -sy(0)-y'(0)\\&=s^2\mathcal{L}[y]-3\end{align}$
$\begin{align}\mathcal{L}[y']&=s\mathcal{L}[y]-y(0)\\&=s\mathcal{L}[y]\end{align}$
Here's where it gets sketchy:
$$\begin{align}\mathcal{L}\left[e^{-x}\sin(x)\right] &= \mathcal{L}\left[e^{-x}\cdot\frac{e^{ix}-e^{-ix}}{2i}\right]\enspace \text{(by Euler's formula)}\\&=\mathcal{L}\left[\frac{e^{x(-1+i)}-e^{x(-1-i)}}{2i}\right]\\&=\frac{1}{2i}\left(\mathcal{L}\left[e^{x(-1+i)}\right]-\mathcal{L}\left[e^{x(-1-i)}\right]\right) \\&=\frac{1}{2i}\left(\frac{1}{s-i+1}-\frac{1}{s+i+1}\right)\end{align}$$
After taking the transform of both sides of the equation and subbing in we have: $$s^2\mathcal{L}[y]-3+2s\mathcal{L}[y]+5\mathcal{L}[y]=\frac{3}{2i}\left(\frac{1}{s-i+1}-\frac{1}{s+i+1}\right)$$ $$\begin{align}&\implies\mathcal{L}[y]\left(s^2+2s+5\right)-3=\frac{3}{2i}\left(\frac{1}{s-i+1}-\frac{1}{s+i+1}\right)\\&\implies\mathcal{L}[y]=\frac{3}{2i(s^2+2s+5)}\cdot\left(\frac{s+i+1-s+i-1}{(s-i+1)(s+i+1)}\right)+3\end{align}\\=\frac{3}{s^2+2s+3}\left(\frac{1}{s^2+s+2}\right)$$ $$\implies y = \mathcal{L^{-1}}\left[\frac{1}{s^4+3s^3+7s^2+7s+6}\right]$$
- Is my reasoning here correct? It's so convoluted that I feel like I've missed an easier way.
- If it is correct, how do I find the inverse transform I've ended up with? I've tried using the tables but I can't figure out which formula to apply.
$$\mathcal{L}\left[e^{-x}\sin(x)\right] = \frac{1}{(s+1)^2+1}$$
1)Take a look at the table here Laplace Transform Table
2) Use fraction decomposition instead of convolution theorem. Or the table. $$g(s)=\frac{3}{s^2+2s+5}$$
$$g(s)=\frac{3}{(s+1)^2+4} \implies g(t)=\frac 32e^{-t}\sin (2t)$$
Decompose the fraction as: $$f(s)= \frac{1}{(s^2+2s+2)(s^2+2s+5)}$$ $$f(s)= \frac{1}{((s+1)^2+1)((s+1)^2+4)}$$ Substitute $u=s+1$ $$f(u)= \frac{1}{(u^2+1)(u^2+4)}$$ $$f(u)= \frac 13\frac{1}{(u^2+1)}- \frac 13\frac{1}{(u^2+4)}$$ $$f(s)= \frac 13\frac{1}{((s+1)^2+1)}- \frac 13\frac{1}{((s+1)^2+4)}$$ Apply inverse Laplace transform. $$f(t)= \frac 13 e^{-t}\sin (t)- \frac 16e^{-t}\sin (2t)$$ Finally $$y(t)=g(t)+3f(t)$$ $$\boxed {y(t)= e^{-t}\sin (t)+ e^{-t}\sin (2t)}$$