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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)]
$\ds\frac{dy}{dx}=x^2y$, where $y=1$ when $x=1$;

\item[(ii)]
$\ds\frac{dy}{dx}-3x^2y=e^{x^3}\cos x$, where $y=1$ when $x=0$.
\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$\ds\frac{dy}{dx}=x^2y$: 1st order, 1st degree, variables
separable

$\Rightarrow  \ds\int\ds\frac{dy}{dx}=\ds\int x^2 \,dx$

$\Rightarrow \ln y=\ds\frac{x^3}{3}+c$

But $y=1$ when $x=1$

$\Rightarrow$

\begin{eqnarray*} \ln 1 & = & \ds\frac{1}{3} +c\\ 0 & = &
\ds\frac{1}{3}+c\\ c & = & -\ds\frac{1}{3} \end{eqnarray*}

$\Rightarrow \ln y=\ds\frac{x^3}{3}-\ds\frac{1}{3}$

or $y=e^{\frac{1}{3}(x^3-1)}$

$e^{-\frac{1}{3}}=0.7165$

\newpage
\item[(ii)]

$\ds\frac{dy}{dx}-3x^2y=e^{x^3}\cos x$: 1st order, 1st degree.

Linear, requiring integrating factor (or exact)

Integrating factor $=e^{\int -3x^2 \,dx}=e^{-x^3}$

Multiply through by it:
$e^{-x^3}\ds\frac{dy}{dx}-3x^2e^{-x^2}y=e^{-x^3}e^{x^3}\cos x$

$\Rightarrow \ds\frac{d}{dx}\left\{ye^{-x^3}\right\}=\cos x$

$\Rightarrow e^{-x^3}y=\ds\int\cos x \,dx+c$

$\Rightarrow y=e^{x^3}[\sin x+c]$

$y=1$ when $x=0$

so $1=e^0[\sin 0+c]$

$\Rightarrow c=1$

Therefore $y=e^{x^3}(\sin x+1)$

\end{description}

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