### Author Topic: TUT0502 QUIZ1  (Read 867 times)

#### Wusijia

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##### TUT0502 QUIZ1
« on: September 27, 2019, 03:08:59 PM »
\begin{equation*}
ty'+(t+1)y=t\ y(ln2)=1\ t>0
\end{equation*}
\begin{equation*}
y'+ \frac{t+1}{t}y=1
\end{equation*}
\begin{equation*}
\mu=e^{\int\frac{t+1}{t}dt}=e^{\int1+\frac{1}{t}dt}=e^{lnt+t}=te^t
\end{equation*}
\begin{equation*}
te^ty'+e^t(t+1)y=te^t
\end{equation*}
\begin{equation*}
(te^ty)'=te^t
\end{equation*}
\begin{equation*}
\int (te^ty)'dt=\int te^tdt
\end{equation*}
\begin{equation*}
{Let} \ u=t \ du=dt \ dv=e^tdt\  v=e^t
\end{equation*}
\begin{equation*}
\int te^tdt=te^t-\int e^tdt=te^t-e^t+C
\end{equation*}
\begin{equation*}
te^ty=te^t-e^t+C
\end{equation*}
\begin{equation*}
y=1-\frac{1}{t}+\frac{C}{te^t}
\end{equation*}
\begin{equation*}
{Since} \ y(ln2)=1
\end{equation*}
\begin{equation*}
1=1-\frac{1}{ln2}+\frac{C}{(ln2)e^{ln2}}
\end{equation*}
\begin{equation*}
\frac{1}{ln2}=\frac{C}{2ln2}
\end{equation*}
\begin{equation*}
C=2
\end{equation*}
\begin{equation*}
{Therefore,}y=1-\frac{1}{t}+\frac{2}{te^t} {,where} \ t>0
\end{equation*}