Determining y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ How do I determine y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ ?

By dividing both sides by $t$ (to obtain the standard form), express the ordinary differential equation as
$$\begin{aligned}
t\dfrac{dy}{dt} + 2t^2y &= 2t^2\\
\dfrac{dy}{dt} + 2ty &= 2t
\end{aligned}$$
Since $p(t) = 2t$ by the standard form of differential equation, the integrating factor of this equation is
$$\begin{aligned}
\mu(t) &= e^{\int 2t\,dt}\\
&= e^{t^2}
\end{aligned}$$
So we have
$$\begin{aligned}
\mu(t)\left(\dfrac{dy}{dt} + 2ty = 2t \right) \longrightarrow e^{t^2}\dfrac{dy}{dt} + 2te^{t^2}y = 2te^{t^2}
\end{aligned}$$
So the general solution is
$$\begin{aligned}
\left[e^{t^2}y\right]' &= 2te^{t^2}\\
e^{t^2}y &= \int 2te^{t^2}\,dt\\
e^{t^2}y &= e^{t^2} + \mbox{C}
\end{aligned}$$
Since $y(1) = 1 + e^3$, we have
$$\begin{aligned}
e(1 + e^3) &= e + \mbox{C}\\
e + e^4 &= e + \mbox{C}\\
\mbox{C} &= e^4
\end{aligned}$$
So this IVP has the solution
$$\begin{aligned}
e^{t^2}y &= e^{t^2} + e^4\\
y(t) &= 1 + e^{4 - t^2}
\end{aligned}$$
Thus,
$$\begin{aligned}
y(2) &= 1 + e^{4 - 2^2}\\
&= 1 + 1 = 2
\end{aligned}$$
