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{\bf Question}

Reduce each of the following equations to normal form and hence
find their general solutions.
\begin{description}
\item[(a)] $y''+{2 \over x}y'+4y=0$
\item[(b)] $y''+2\sin x\,y'+\cos x(1-\cos x)y=0$
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds y''+{2 \over x}y'+4y=0$. So $P(x)={2 \over x}$ and $Q(x)=4$.

We transform to a new variable $u(x)$ where $y(x)=u(x)v(x)$ and

$\ds v(x)=\exp\left(-{1 \over 2}\int Pdx\right)=\exp\int-{{dx} \over x}
=\exp(-\ln x)={1 \over x}$.

So $y(x)={{u(x)} \over x}$.

The equation for $u(x)$ is $u''+q(x)u=0$ where $q=Q-{1 \over 4}P^2-{1
\over 2}P'$

Substituting for $P$ and $Q$ we find $q=4-{1 \over {x^2}}+{1 \over
{x^2}}=4$.

So that $u''+4u=0, \quad \Rightarrow \quad u(x)=A\cos2x+B\sin2x$.

Hence $y(x)={1 \over x}(A\cos2x+B\sin2x)$.

\item[(b)]
$\ds y''+2\sin xy'+\cos x(1-\cos x)y$.

So $P(x)=2\sin x$ and $Q(x)=\cos x(1-\cos x)$.

We transform to a new variable $u(x)$ where $y(x)=u(x)v(x)$ and

$\ds v(x)=\exp\left(-{1 \over 2}\int Pdx\right)=\exp\int-\sin x dx
=e^{\cos x}$.

So $y(x)=e^{\cos x}u(x)$.

The equation for $u(x)$ is $u''+q(x)u=0$ where $q=Q-{1 \over 4}P^2-{1
\over 2}P'$

Substituting for $P$ and $Q$ we find $q=\cos x-\cos^2x-\sin^2x-\cos x=-1$.

So that $u''-u=0, \quad \Rightarrow \quad u(x)=Ae^x+Be^{-x}$.

Hence $y(x)=e^{\cos x}(Ae^x+Be^{-x})$.
\end{description}



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