The differential equation solution with Laplace transform

Question

I am stuck with this equation. If you can help me

$ y''(t) + 12 y'(t) + 32 y(t) = 32 u(t)$ with $y(0) = y'(0) = 0$

I found the laplace transform for $y(t) $

$Y(p) = x = \frac{32}{(p(p^{2}+ 12p+32))} $

so i need the laplace transform of $y(t)$ and then the solution for $y(t)$. thanks in advance ps: this is my first post here, so i don't know the protocol. :)

Answer 1

$$\frac{32}{p(p^2+12p+32)} = \frac{32}{p((p+6)^2-2^2)} = \frac{32}{p(p+4)(p+8)}$$

Now use partial fractions method to get $$\frac{a}{p}+\frac{b}{p+4}+\frac{c}{p+8}$$

Then the Laplace inverse would be,

$$a + b^{-4t} + c^{-8t}$$