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QUESTION


\begin{description}

\item[(i)]
Find the solution to the following differential equation given the
condition $y(0)=0$:

$$\frac{dy}{dx}=\sin x\cos^2y.$$

\item[(ii)]
Find the general solution to the differential equation. Note any
values of $x$ for which the solutions may not be valid:

$$x\log x\frac{dy}{dx}+y=2\log x.$$

[Hint: Note that $\frac{d}{dx}\{\log(\log x)\}=\frac{1}{x\log
x}$.]

\item[(iii)]
Find the solution to the following differential equation given the
conditions: $y(0)=1,y'(0)=7$:

$$\frac{d^2y}{dx^2}-10\frac{dy}{dx}+29y=0.$$

\end{description}



ANSWER


\begin{description}

\item[(i)]
Separable

\begin{eqnarray*}
\int\frac{1}{\cos^2y}\,dy&=&\int\sin x\,dx\\ \tan y&=&-\cos x+k\\
y(0)=0&\Rightarrow&\tan(0)=-\cos(0)+k\\ &\Rightarrow&0=k-1\textrm{
or }k=1\\ \tan y&=&1-\cos x
\end{eqnarray*}

\item[(ii)]
Integrating factor

$$\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2\log x}{x\log x}\
x\neq0,1$$

$$IF=e^{\int\frac{1}{x\log x}\,dx}=e^{\log(\log x)}=\log x$$

so

$$y\log x=\int\frac{2\log x}{x}\,dx=\log^2x+k$$

by putting $u=\log x$

or

$$y=\frac{k}{\log x}+\log x,\ x\neq1$$

This is valid for $x>o,\ x\neq1$

\item[(iii)]

The auxilliary equation is $\lambda^2-10\lambda+29=0$ which has
roots $\lambda=\frac{10\pm\sqrt{100-116}}{2}=5\pm2i$

The solution is therefore

$$y=e^{5x}(A\cos2x+B\sin2x)$$

$1=y(0)=A,$

$y'(0)=5e^{5x}(A\cos2x+B\sin2x)+e^{5x}(-2A\sin 2x+2B\cos2x)$

so $7=y'(0)=5A+2B=5+2B\Rightarrow B=1$

Hence the solution is

$$y=e^{5x}(\cos 2x+\sin 2x)$$

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