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QUESTION


\begin{description}

\item[(a)]
Find the solution to the following differential equation:

$$(x^3-7)\frac{dy}{dx}+3x^2y^2=0,$$

subject to the initial condition $y=1$ when $x=2$.

\item[(b)]
Find the solution to the following differential equation

$$x\frac{dy}{dx}+2y=\sin(x),$$

subject to the initial condition $y=\frac{4}{\pi^2}$ when
$x=\frac{\pi}{2}$.

\item[(c)]
Fins the solution to the following differential equation given the
conditions:

$$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+5y=0,$$

subject to the initial conditions

$y(0)=1,\frac{dy}{dx}(0)=5$.

\end{description}



ANSWER


\begin{description}

\item[(a)]
Rearranging we get $\ds\frac{dy}{dx}=\frac{-3x^2y^2}{(x^3-7)}$,
which is separable with
$\ds\frac{1}{y^2}\frac{dy}{dx}=\frac{-3x^2}{(x^3-7)}$ giving
$\ds\frac{-1}{y}=\int\frac{-3x^2}{x^3-7}\,dx$. Now put $u=x^3-7$
to get $\ds\frac{du}{dx}=3x^2$ so that
$\ds\int\frac{-3x^2}{x^3-7}\,dx=\int\frac{\frac{-du}{dx}}{u}\,dx=
\int\frac{-1}{u}\,du=-\log|u|+k-\log|x^3-7|+k$. Hence $\ds
y=\frac{1}{\log|x^3-7|-k}$. Putting $y=1$ when $x=2$ gives
$\ds1=\frac{1}{-k}$ so $k=-1$ and $\ds y=\frac{1}{\log|x^3-7|+1}$.

\item[(b)]
Assuming $x\neq0$ simplify to get
$\ds\frac{dy}{dx}+\frac{2}{x}y=\frac{\sin(x)}{x}$ which is first
order linear with integrating factor $\ds
e^{\int\frac{2}{x}\,dx}=x^2$. The equation becomes $\ds x^2y=\int
x\sin x\,dx$. Integrate by parts with $u=x$ and
$\ds\frac{dv}{dx}=\sin x$ to get $x^2y=[-x\cos x]+\sin x+k$, so
$\ds y=\frac{-\cos x}{x}+\frac{\sin x}{x^2}+\frac{k}{x^2}$.
Putting $\ds y=\frac{4}{\pi^2}$ when $\ds x=\frac{\pi}{2}$ we get
$k=0$ so $\ds y=\frac{-\cos x}{x}+\frac{\sin x}{x^2}$.

\item[(c)]
Second order homogeneous linear with auxilliary equation
$\ds\lambda^2-2\lambda+5-0$. Roots $1\pm2i$ so general solution
has the form $\ds y=e^x[A\cos2x+B\sin2x]$. Then
\begin{eqnarray*}
\frac{dy}{dx}&amp;=&amp;e^x[-2A\sin2x+2B\cos2x]+e^x[A\cos 2x+B\sin 2x]\\
&amp;=&amp; e^x[(2B+A)\cos x+(B-2A)\sin x]
\end{eqnarray*}

Putting $y(0)=1$, and $\ds\frac{dy}{dx}(0)=5$ we get $\ds A+2B=5\
A=1\Rightarrow B=2$ so $\ds y=e^x[\cos2x+2\sin2x]$.


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