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{\bf Question}

State the order and degree of the following differential
equations, identify their type and hence solve them.

\begin{description}
\item[(i)]
$\ds\frac{dy}{dx}=xy$, where $y=3$ when $x=0$;

\item[(ii)]
$\ds\frac{dy}{dx}-2xy=\ds\frac{e^{x^2}}{1+x}$, where $y=0$ when
$x=0$.
\end{description}
\medskip

{\bf Answer}

Both are first order first degree.

\begin{description}
\item[(i)]
This is V.S.

\begin{eqnarray*} & & \ds\frac{dy}{dx}=xy\\ & \Rightarrow &
\ds\int\ds\frac{dy}{y}=\ds\int x \,dx\\ & \Rightarrow & \ln
y=\ds\frac{x^2}{2}+C\\ & \Rightarrow & y=Ae^{\frac{x^2}{2}}\\ y=3\
\rm{when}\ x=0\\ & \Rightarrow & 3=Ae^0=A \end{eqnarray*}

Therefore \un{$y=3e^{\frac{x^2}{2}}$}

\item[(ii)]
This is \un{linear}

$$\ds\frac{dy}{dx}-2xy=\ds\frac{e^{x^2}}{1+x}$$

Integrating factor $=e^{\int -2x\, dx}=e^{-x^2}$

Therefore
$e^{-x^2}\ds\frac{dy}{dx}-2xe^{-x^2}y=e^{-x^2}\ds\frac{e^{-x^2}}{1+x}$

$\Rightarrow
\ds\frac{d}{dx}\left\{ye^{-x^2}\right\}=\ds\frac{1}{1+x}$

Therefore \begin{eqnarray*} ye^{-x^2}=\ds\int\ds\frac{dx}{1+x}\\ &
= & \log(1+x)+c \end{eqnarray*}

Therefore $y=e^{x^2}[\log(1+x)+c]$

$y=0$ when $x=0$

$\Rightarrow \begin{array}{rcl} 0 & = & e^0[\log 1+c]\\ 0 & = & c
\end{array}$

Therefore \un{$y=e^{x^2}\log(1+x)$}

\end{description}
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