Answer
the solution is
$$
y(t)=[\frac{1}{3}t^{3}+c] e^{2t}
$$
where $c$ is a constant
Work Step by Step
$y^{'} -2y=t^{2 }e^{2t}$
we know that $p(t)=-2$ , $g(t)=t^{2}e^{2t}$ so the homogeneous solution is
$$y(t)=A e^{-\int{p(t)dt}}=A e^{2t}$$
where $A$ is a constant.
By using the method of variation of parameters we assume that the solution is in the form
$$y(t)=A(t) e^{-\int{p(t)dt}}=A(t) e^{2t} (*) $$
where $A(t)$ is now a function of t
A(t) must satisfy the condition
$$A^{'}(t)= g(t)e^{\int{p(t)dt}}=t^{2}e^{2t}. e^{-2t}=t^{2}$$
by integration we get :
$$A(t)= g(t){\int{t^{2}dt}}= \frac{1}{3}t^{3}+c$$
where $c$ is a constant
Then substitute for A(t) in Eq. (*) and determine y.
the solution is
$$
y(t)=[\frac{1}{3}t^{3}+c] e^{2t}
$$
where $c$ is a constant
