Use Laplace transform to solve the following initial–value problems. Use Laplace transform to solve the following initial–value problem.
$y′′′′ + 2y′′ + y = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0, y′′′(0) = 2$
Answer
$s^4 L(s) - s^3y(0) -s^2 y'(0) - s y''(0) - y'''(0) +2[s^2L(s)-sy(0)-y'(0)] +L(s) \\\\$
I get the partial fraction part and got stuck, need help!
$L(s) =\frac{s^3 - s^2 +2s}{s^4 +2s^2 +1}= \frac{s-1}{s^2 +1}+\frac{s+1}{(s^2+1)^2}  \:\:$Factorising the denominator I get: $(s^2+1)^2$
Please some let me know if Im heading in the wrong direction here.

I guess it is the last fraction, $\frac{1+s}{(s^2+1)^2}$, that causes you problems.
For the (one sided) Laplace transform we have the transform pairs
$$\sin{t}\stackrel{\mathcal{L}}{\longmapsto}\frac{1}{s^2+1},$$
$$tf(t)\stackrel{\mathcal{L}}{\longmapsto}-F'(s),$$
and
$$f'(t)\stackrel{\mathcal{L}}{\longmapsto}sF(s).$$
The first two mean that
$$t\sin{t}\stackrel{\mathcal{L}}{\longmapsto}\frac{2s}{(s^2+1)^2},$$
and the third rule gives a hint on how to get rid of the $s$ in the numerator.
