Author Topic: TUT5103 quiz1  (Read 889 times)

hz12

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TUT5103 quiz1
« on: September 26, 2019, 11:50:11 PM »
\begin{aligned}
t y^{\prime}+2 y&=\sin (t), t>0\\
y^{\prime}+\frac{2}{t} y&=\frac{\sin (t)}{t}\\
y^{\prime}+p(t) y&=g(t)
\end{aligned}
$$
\therefore P(t)=\frac{2}{t} \quad g(t)=\frac{\sin (t)}{t}
$$
$$
\mu(t)=e^{\int \frac{2}{t} d t}=e^{2 \ln (t)}=t^{2}
$$
Muttiply $\mu(t)$ to standard equation
$$
\begin{array}{c}{t^{2} y^{\prime}+2 t y=t \sin t} \\ {\mathrm{and}\left(t^{2} y\right)^{\prime}=t \sin t}\end{array}
$$
Integrating  both  sides:
$$
\int\left(t^{2} y\right)^{\prime}=\int t \sin t
$$
Thus, $$t^{2} y=\int t \sin t$$
Integrating by parts:
$$
\begin{aligned}
u&=t, \quad d V=\sin t\\
d u&=d t \quad V=-\cos t
\end{aligned}
$$
$$
\begin{aligned}
\therefore \int t \sin t &=u v-\int v d u \\
 &=-t \cos t-\int-\cos t d t \\
  &=-t \cos t+\int \cos t d t \\
   &=-t\cos t+\sin t+C \\
   \end{aligned}

$$
\begin{aligned}
\therefore t^{2} y&=-t \cos t+\sin t+C\\
y&=\displaystyle\frac{-t \cos t+\sin t+C}{t^{2}}
\end{aligned}

since given $t>0$, $y \rightarrow 0$ as $t \rightarrow \infty$









« Last Edit: September 27, 2019, 01:10:04 PM by Hanyu Zhou »

Xinqiao Li

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Re: TUT5103 quiz1
« Reply #1 on: September 27, 2019, 10:45:42 AM »
a very small typo: the final solution should be just y, not y^2.