Let x(t) be the solution of the initial-value problem(a) Find the Laplace transform F(s) of the forcing f(t).(b) Find the Laplace transform X(s) of the solution x(t).x"+8x'+20x=f(t)x(0)=-3x'(0)

given differential equation is x
$x"+8x^{\prime }+20x=f(t)$
$x(0)=-3$
$x^{\prime }(0)=5$
$\text{where the forcing }f(t)\text{ is given by }$
$f(t)=\{\begin{array}{ll}{t}^2& \text{for }0\le t<2,\\ 4e^{2-t}& \text{for }2\le t<\mathrm{\infty }.\end{array}$
$\therefore x"+8x^{\prime }+20x=t^2+4e^{2-t}$
$\text{we know that }L\left[x"\right]=s^{2}X(s)-sX(0)-x^{\prime }(0)$
$L\left[x^{\prime }\right]=sX(s)-x(0)$
$L\left[t^n\right]=\frac{n!}{s^{n+1}},L\left\{e^{-at}\right\}=\frac{1}{s+a}$
$\text{Step 2}$
$\text{Taking Laplace tranform of equation}$
$L\left[x"\right]+8L\left[x^{\prime }\right]+20L\left[x\right]=L\left[t^2\right]+4e^{2}L\left\{e^{-t}\right\}$
$s^2(X(s))-sX(0)-x^{\prime }(0)+8[s(X(s))-x(0)]+20X(s)=\frac{2!}{s^{2+1}}+\frac{4e^2}{s+1}$
$\therefore s^{2}X(s)-s(-3)-5+8(sX(s)-5)+20X(s)=\frac{2}{s^3}+\frac{4e^2}{s+1}$
$\therefore (s^2+8s+20)X(s)+3s-5-40=\frac{2}{s^3}+\frac{4e^2}{s+1}$
$\therefore (s^2+8s+20)X(s)+3s-45=\frac{2}{s^3}+\frac{4e^2}{s+1}$
$\therefore X(s)[s^2+8s+20]=\frac{2}{s^3}+\frac{4e^2}{s+1}+45-3s$
$\therefore x(s)=\frac{1}{s^2+8s+20}\{(45-3s)+\frac{2}{s^3}+\frac{4e^2}{s+1}\}$