Solution of differential equation using Laplace transform

Question

Can someone please help me identify my mistake as I am not able to find it.

$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$

$L[y'']+L[y']=3L[x^2]$

$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$

$L[y']=pL[y]-y(0)=pL[y]$

$p^2L[y]-1+pL[y]=\frac{6}{p^3}$

$p^2L[y]+pL[y]=\frac{6}{p^3}+1$

$(p^2+p)L[y]=\frac{6}{p^3}+1$

$L[y]=\frac{6}{p^3(p^2+p)}+\frac{1}{p^2+p}$

$y=x^3-e^{-x}+1-e^{-x}$

Answer 1

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

Answer 2

Using \begin{align} \mathcal{L}\{y'\} &= s \, \overline{y} - y(0) \\ \mathcal{L}\{y''\} &= s^2 \, \overline{y} - s \, y(0) - y'(0) \end{align} then $$y'' + y' = a \, t^2$$ transforms to \begin{align} s \, (s+1) \, \overline{y} &= s \, y(0) + y'(0) + y(0) + \frac{2 a}{s^3} \\ \overline{y} &= \frac{y(0) + y'(0)}{s \, (s+1)} + \frac{y(0)}{s+1} + \frac{2 a}{s^4 \, (s+1)} \\ &= \frac{y(0) + y'(0) - 2 a}{s} + \frac{2 a - y'(0)}{s+1} + \frac{2 a}{s^4} - \frac{2 a}{s^3} + \frac{2 a}{s^2} \\ &= \mathcal{L}\{ (2a - y'(0)) \, e^{-t} + \frac{a}{3} \, t^3 - a \, t^2 + 2 a \, t + y(0) + y'(0) - 2 a \}. \end{align}

This leads to the general solution $$y(t) = (2a - y'(0)) \, e^{-t} + \frac{a}{3} \, t^3 - a \, t^2 + 2 a \, t + y(0) + y'(0) - 2 a. $$ Now applying the conditions $y(0) = 0$, $y'(0) =1$, $a=3$ then $$y(t) = 5 \, e^{-t} + t^3 - 3 \, t^2 + 6 \, t - 5. $$

Note that a key component is $$\frac{1}{s^4 \, (s+1)} = \frac{1}{s+1} + \frac{1}{s^4} - \frac{1}{s^3} + \frac{1}{s^2} - \frac{1}{s}.$$