Using Laplace Transform to solve $y'' + 4y' + 4y =0, \quad y(0)=1, y'(0)=1$

Question

These were the steps I took:

$sL[y'] - y'(0) + 4sL[y] - 4y(0) + 4L[y] = 0$

$s\big(sL[y] - y(0)\big) - 1 + 4sL[y] - 4 + 4L[y] = 0$

$s^2L[y] - s - 1 + 4sL[y] - 4 + 4L[y] = 0$

$s^2L[y] - s + 4sL[y] + 4L[y] -5 = 0$

$(s^2 + 4s + 4)L[y] = 5 + s$

$L[y] = \frac{5 + s}{s^2 + 4s + 4}$

$L[y] = \frac{5 + s}{(s + 2)^2}$

I got stuck here

But the final answer was $ y = (1+3x)e^{-2x}$

Please how do I get to this answer

Answer 1

Hint:$$\frac{5+s}{(s+2)^2} = \frac{2+s+3}{(s+2)^2} = \frac1{s+2}+\frac3{(s+2)^2}$$

Then use the fact that $\mathcal{L}^{-1}\{s+a\} = e^{-ax}\mathcal{L}^{-1}\{s\}$