Solving a differential equation by Laplace transform

Question

I am trying to solve:

$$y"+2y'+y = 0, y(0)=1,y'(0)=1$$

My work is shown below:

$y''+2y'+y=0$, $y(0)=1$, $y'(0)=1$

$\mathcal L[y''+2y'+y]=\mathcal L(0)$

$\mathcal L[y'']+2\mathcal L[y]'+\mathcal L[y]=0$

$[s^2\mathcal L[y(s)]-sy(0)-y'(0)]+2[sx(s)-y(0)]+x(s)=0$

$s^2x(s)-s-1+2sx(s)-2+x(s)=0$

$x(s)(s^2+2s+1)=s+3$

$x(s)=\frac{s+3}{s^2+2s+1}$

I don't see this particular problem online or any other site so it might help people on the same problem.

Any advice you might have as to how to proceed would be greatly appreciated. Thanks in advance.

Answer 1

You can use this $$\mathcal{L}(e^{at}t^n)=\frac {n!} {(s-a)^{n+1}}$$ And also $$\frac {s+3}{(s+1)^2}=\frac 1 {s+1}+\frac 2{(s+1)^2}$$

Answer 2

Thanks for the help, is the answer correct?

Answer is :

$$y(t)= e^{-t} + 2te^{-t}$$ not the one on the paper