# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: Tiantian Yu on September 27, 2019, 07:03:14 PM

Title: TUT0303 Quiz1
Post by: Tiantian Yu on September 27, 2019, 07:03:14 PM
Find the general solution of
\begin{equation*}ty'-y=t^{2}e^{-t}
\end{equation*}
and use it to determine how solutions behave as t → ∞.

put the differential equation into standard form: \begin{equation*}y'-\frac{y}{t}=\frac{t^{2}e^{-t}}{t}\end{equation*}
then \begin{equation*}p(t)=-\frac{1}{t}\end{equation*}
the integrating factor is \begin{equation*}μ(t)= e^{\int-\frac{1}{t}dt}=e^{-ln|t|} =\frac{1}{|t|}\end{equation*}
since t>0,we take\begin{equation*}\frac{1}{t}\end{equation*}
\begin{equation*}\frac{d}{dx}(\frac{1}{t}y)=\frac{t^{2}e^{-t}}{t}\frac{1}{t}\end{equation*}
integrating both sides, we have\begin{equation*}\frac{1}{t}y=\int\frac{t^{2}e^{-t}}{t^{2}}dt=-e^{-t}+C\end{equation*}
\begin{equation*}y=-te^{-t}+Ct\end{equation*}
the behaviour of the solution is determined by the term Ct
if c=0,\begin{equation*}\lim_{t\to\infty} -te^{-t}=-\lim_{t\to\infty} te^{-t}=0\end{equation*}
if c is positive,\begin{equation*}\lim_{t\to\infty} -te^{-t}+Ct=\lim_{t\to\infty} Ct=\infty\end{equation*}
if c is negative,\begin{equation*}\lim_{t\to\infty} -te^{-t}-Ct=-\lim_{t\to\infty} Ct=-\infty\end{equation*}
Thus the general solution of the given differential equation is\begin{equation*}y=-te^{-t}+Ct\end{equation*}and y=∞,-∞, or 0 as t->∞