Using laplace transform, find the solution $y=y(x)$ to the equation $2y'''(x)+y''(x)-y'(x)=3\sin x$ that satisfies the conditions $y(0)=1$ $y'(0)=0$ and $y''(0)=0$
I have taken a laplace transform (for now just showing the l.h.s:
$2\mathcal{L}(y''')+\mathcal{L}(y'')-\mathcal{L}(y') = -y(0)+2p\mathcal{L}(y'')+p\mathcal{L}(y')-p\mathcal{L}(y)$
$=-y'(0)-py(0)+2p^2\mathcal{L}(y')+p^2\mathcal{L}(y)-p\mathcal{L}(y)$
$=-y''(0)-py'(0)-p^2y(0)+2p3\mathcal{L}(y)+p^2\mathcal{L}(y)-p\mathcal{L}(y$
This would give me: $-p^2+(2p^3+p^2-p)\mathcal{L}(y)$
However my answer book gives: $-2p^2-p+1+(2p^3+p^2-p)\mathcal{L}(y)$
Is there an error in the book or if not could someone correct my workings please