Cannot solve a differential equation using Laplace transformations

Question

$\dddot x + x = 2t$, $x(0)=0$, $\dot x(0)=1$, $\ddot x(0)=0$ and the answer should be $x=t^2-2+2\cos{t}+\sin{t}$

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I start off with assuming $L(x)=X(p)$ then $$\dddot x=p^3X(p)-p^2X(0)-pX'(0)-X''(0)=p^3X-p$$ $$p^3X-p+X=\frac2{p^2}$$ $$X=\frac2{p^2(p^3+1)}+\frac{p}{p^3+1}=2\frac1{p^2}+\frac13\frac1{p+1}-\frac13\frac{p+1}{p^2-p+1}$$ $$\frac{p+1}{p^2-p+1}=\frac{p-\frac12+\frac12+1}{p^2-p+\frac14+\frac34}=\frac{p-\frac12}{(p-\frac12)^2+\frac34}+\sqrt{3}\frac{\frac{\sqrt3}{2}}{(p-\frac12)^2+\frac34}$$ $$X=2t +\frac13e^{-t}-\frac13e^{\frac{t}2}cos{\frac{\sqrt3}2t}-\frac{\sqrt3}3e^{\frac{t}2}sin{\frac{\sqrt3}2t}$$

Answer 1

If $\dddot x+\dot x=2t, x(0)=\ddot x(0)=0, \dot x(0)=1$ $$p^3X-p+pX=\frac2{p^2}$$ $$X=\frac{p^3+2}{p^3(p^2+1)}=\frac{A}{p^3}+\frac{B}{p^2}+\frac{C}{p}+\frac{Dp+E}{p^2+1}$$ $$A=2, B=0, C=-2, D=2, E=1$$ $$X=2\frac1{p^2}-2\frac1{p}+\frac{2p+1}{p^2+1}=2\frac1{p^2}-2\frac1{p}+2\frac{p}{p^2+1}+\frac1{p^2+1}=2t^2-2+2cost+sint$$ which matches the answer I've been given. Thanks to @Lutz Lehmann for pointing it out