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\noindent Find the general solution of the given differential equation, and use it to determine how solutions
behave as $t \rightarrow \infty$
$$
t y^{\prime}-y=t^{2} e^{-t}, \quad t>0
$$
$$
\begin{array}{l}{y^{\prime}-\frac{1}{t} y=t e^{-t}} \\ {p(t)=-\frac{1}{t}} \\ {\mu=e^{\int p(t) d t}=e^{-\ln (t)}=t^{-1}}\end{array}
$$
multiply both sides by $\mu$
$$
\begin{aligned} t^{-1} y^{\prime}\cdot t^{-2} y &=e^{-t} \\ \frac{d}{d x}\left(t^{-1} y\right) &=e^{-t} \\ t^{-1} y &=\int e^{-t} \\ t^{-1} y &=-e^{-t}+c \\ y &=-t e^{-t} + c t \end{aligned}
$$
when $t \rightarrow \infty$
case 1: $$
\begin{array}{l}{C=0;} \\ {\text { by } L^{\prime} \text { Hopital rule; }} \\ {y \rightarrow 0}\end{array}
$$
case 2:$$
\begin{array}{l}{C>0;} \\ {y \rightarrow+\infty}\end{array}
$$
case 3:$$
\begin{array}{l}{C<0;} \\ {y \rightarrow-\infty}\end{array}
$$
