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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}
Find the general solution of the following differential equations

\begin{description}
\item[(i)]
$\ds\frac{dy}{dx}=\sin x$

\item[(ii)]
$e^x\ds\frac{dy}{dx}=\sin x$

\item[(iii)]
$\ds\frac{dy}{dx}=1+y$

\item[(iv)]
$\ds\frac{dy}{dx}=1+y^2$

\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
${}$

$\ds\frac{dy}{dx}=\sin x$ is of the form$\ds\frac{dy}{dx}=f(x)$,
type (i).

Solve by direct integration:

$\begin{array} {l} \Rightarrow \ds\int\,dy=\ds\int\sin x \,dx\\
\Rightarrow \un{y=-\cos x+c} \end{array}$

\item[(ii)]
${}$

$e^x \ds\frac{dy}{dx}=\sin x$.

Rearrange to get $\ds\frac{dy}{d}=e^{-x}\sin x$. Again of the form
$\ds\frac{dy}{dx}=f(x)$, type (i).

Thus

$\ds\int dy = \ds\int e^{-x} \sin x dx$

integrate by parts with $\left\{\begin{array} {ll} u=e^{-x} &amp;
\frac{dv}{dx}=\sin x\\ \frac{du}{dx}=-e^{-x} &amp; v=-\cos x
\end{array}\right.$

$\Rightarrow y=-e^{-x}\cos x-\ds\int e^{-x} \cos x dx$

integrate by parts with $\left\{\begin{array} {ll} u=e^{-x} &amp;
\frac{dv}{dx}=\cos x\\ \frac{du}{dx}=-e^{-x} &amp; v=\sin x
\end{array}\right.$

$\Rightarrow y=-e^{-x}\cos x-[e^{-x} \sin x]+\ds\int -e^{-x}\sin x
dx$

$\Rightarrow y=-e^{-x}(\cos x+\sin x)-y+c$

$\Rightarrow \un{y=-\ds\frac{e^{-x}}{2}(\cos x+\sin x)+c'}$

\item[(iii)]
${}$

$\ds\frac{dy}{dx}=1+y$ is of the form $\ds\frac{dy}{dx}=f(y)$ so
rearrange to get

$$\begin{array} {rcl} &amp; &amp; \ds\int\ds\frac{dy}{1+y}=\ds\int dx\\ &amp;
\Rightarrow &amp; \ln(1+y)=x+c\\ &amp; \Rightarrow &amp; 1+y=e^{x+c}\\ &amp;
\Rightarrow &amp; \un{y=e^{x+c}-1} \end{array}$$

\newpage
\item[(iv)]
${}$

$\ds\frac{dy}{dx}=1+y^2$ is of the form $\ds\frac{dy}{dx}=f9y)$ so
rearrange to get

$$\begin{array} {rcl} &amp; &amp; \ds\int\ds\frac{dy}{1+y^2}=\ds\int dx\\
&amp; \Rightarrow &amp; \arctan y=x+c\\ &amp; \Rightarrow &amp; \un{y=\tan(x+c)}
\end{array}$$
\end{description}
\end{document}
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