### Author Topic: TUT0402 quiz1  (Read 494 times)

#### Junhong Zhou

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##### TUT0402 quiz1
« on: September 27, 2019, 02:00:07 PM »
$$ty'+2y=sin(t)$$

write this into standard equation:
$$y'+\frac{2}{t}y=\frac{sin(t)}{t}$$

which here we have
$$p(t)=\frac{2}{t}$$
$$\mu(x)=e^{\int p(t)dt}=e^{\int\frac{2}{t}dt}=e^{2\ln{t}}=t^2$$

multiply $\mu(x)$ to both sides of the standard equation we have:
\begin{align}
(t^2)y'+(t^2)\frac{2}{t}y &= (t^2)\frac{sin(t)}{t}\notag\\
t^2y'+2ty &= tsin(t)\notag\\
\int t^2y'+2tydt &=\int tsin(t)dt\notag\\
t^2y &= \int tsin(t) \notag
\end{align}

integrate $\int tsin(t)dt$ by parts, let $u=t$ and $dv=sin(t)dt$:
\begin{align}
\int tsin(t)dt &= t(-cos(t))-\int -cos(t)dt\notag\\
&= -tcos(t)+\int cos(t)dt\notag\\
&= -tcos(t)+sin(t)+C\notag
\end{align}

so we have:
\begin{align}
t^2y &= -tcos(t)+sin(t)+C\notag\\
y &= \frac{-tcos(t)+sin(t)+C}{t^2}\notag
\end{align}

So:
$$\lim_{t\to\infty}y=\lim_{t\to\infty}\frac{-tcos(t)+sin(t)+C}{t^2}=0$$