Answer
The solution is :
$$y(t)=3[t^{2}-4t+8]+c e^{\frac{-t}{2}}
$$
where $c$ is a constant.
Work Step by Step
$$2y^{'}+y=3t^{2}, $$
dividing by $2$ we have :
$$y^{'}+\frac{1}{2}y=\frac{3}{2}t^{2}, \quad \quad (* )$$
we know that :
$$p(t)=\frac{1}{2}, g(t)=\frac{3}{2}t^{2}$$
The homogeneous solution is
$$y(t)=Ae^{\int{-p(t)dt}}=Ae^{\int{-\frac{1}{2}dt}}=Ae^{-\frac{t}{2}}$$
where $A$ is a constant.
By using the method of variation of parameters, we assume that the solution of Eq. (*) is of the form
$$ y(t)=A(t)e^{-\frac{t}{2}}, \quad \quad (**)
$$
where A is now a function of t. By substituting for $y(t)$ in the given differential equation (*), show that A(t) must satisfy the condition
$$A^{'}(t)=g(t)e^{\int{p(t)dt}}=\frac{3}{2} t^{2}e^{\frac{t}{2}}
$$
by integration we find
$$A(t)=\frac{3}{2}\int { t^{2}e^{\frac{t}{2}}dt }
$$
integration by parts two times shows that :
$$A(t)=3e^{\frac{t}{2}}[t^{2}-4t+8]+c
$$
where $c$ is a constant.
Then substitute for A(t) in Eq. (**) and determine $y(t)$.
$$y(t)=3[t^{2}-4t+8]+c e^{\frac{-t}{2}}
$$
