Solution of differential equation using Laplace transform Can someone please help me identify my mistake as I am not able to find it.
$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$
$L[y'']+L[y']=3L[x^2]$
$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$
$L[y']=pL[y]-y(0)=pL[y]$
$p^2L[y]-1+pL[y]=\frac{6}{p^3}$
$p^2L[y]+pL[y]=\frac{6}{p^3}+1$
$(p^2+p)L[y]=\frac{6}{p^3}+1$
$L[y]=\frac{6}{p^3(p^2+p)}+\frac{1}{p^2+p}$
$y=x^3-e^{-x}+1-e^{-x}$

$\frac{6}{p^3(p^2+p)}=\frac{6}{p^4} - \frac{6}{p^3} + \frac{6}{p^2} - \frac{6}{p} + \frac{6}{1+p}$ then reverse transform gives you $x^3-3x^2+6x-6+6e^{-x}$ add the rest which is right in your calculation we get $$y=x^3-3x^2+6x-6+6e^{-x}+1-e^{-x}=x^3-3x^2+6x-5+5e^{-x}$$
