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.

Great job getting to this point, now we just have to get the answer in forms we can work with or use the formal definitions for inverse Laplace transforms. I will use known forms.
We have (split up the numerators):
$$\dfrac{s-1}{s^2 +1}+\dfrac{s+1}{(s^2+1)^2} = \dfrac{s}{s^2+1} - \dfrac{1}{s^2+1} + \dfrac{s}{(s^2+1)^2} + \dfrac{1}{(s^2+1)^2}$$
The inverse LT of this (using Laplace tables) is given by:
$$y(x) = \cos x - \sin x + \dfrac{1}{2} x \sin x + \dfrac{1}{2} (\sin x -x \cos x)$$
So, some simple algebra yields:
$$y(x) = \cos x - \dfrac{1}{2}\sin x + \dfrac{1}{2} x \sin x - \dfrac{1}{2} x \cos x$$
You should verify this result satisfies the original ODE.
