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\begin{document}


{\bf Question}

Use the method of reduction of order to obtain a second solution
to each of the following equations using the given solution $y_1$
\begin{description}
\item[(a)] $y''-4y'+5y=0, \qquad y_1=e^{2x}\sin x$,
\item[(b)]$(x^2+1)y''+2xy'=0, \qquad y_1=c$, a constant,
\item[(c)]$x^3y'''+3x^2y''+xy'-8y=0, \qquad y_1=x^2$.
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)] $y''-4y'+5y=0, \qquad y_1=e^{2x}\sin x$

$\ds P(x) = -4$ so $\ds \exp \int -P dx = e^{4x}$

and $\ds\frac{1}{y_1^2} \exp \int -P dx = \frac{1}{\sin^2x} $

\begin{eqnarray*}
v &amp; = &amp; \int \frac{1}{e^{4x}\sin^2x} \cdot e^{4x} \, dx \\ &amp; = &amp;
\int \frac{dx}{\sin^2 x} \\ &amp; = &amp;  \csc^2x \, dx \\ &amp; =&amp; -\cot x
\end{eqnarray*}

Therefore $y_2 = vy_1 = -e^{2x} \cos x$ is a second solution.

The General solution is $$y = -e^{2x}(A \cos x + B \sin x).$$

\item[(b)]$(x^2+1)y''+2xy'=0, \qquad y_1=c$, a constant,

Therefore $\ds P(x) = \frac{2x}{x^2 +1}$ so \begin{eqnarray*} \exp
\int -P \, dx &amp; =&amp; \exp \int -\frac{2x}{x^2+1}dx\\ &amp; = &amp;
\exp(-\ln(x^2+1)) \\ &amp; = &amp;\frac{1}{x^2+1}\end{eqnarray*} Thus $\ds
\frac{1}{y_1^2} \exp \int -P \, dx = \frac{1}{x^2+1}$ taking (c =
1)

and $\ds v = \int \frac{dx}{x^2+1} = \arctan x$

So $y_2 = vy_1 = \arctan x$ is a second solution.

The general solution is $$ y = A + B \, \arctan x $$
\item[(c)]
$x^3y'''+3x^2y''+xy'-8y=0 \hspace{.7in} (1)$

Let $y_1 = x^2 {\rm \ \ and \ \ \ } y = vy_1 = x^2v$
\begin{eqnarray*} y'&amp; = &amp; 2xv + x^2v'
\\ y'' &amp; = &amp; 2v + 4xv' + x^2v'' \\ y''' &amp; = &amp; 6v' + 6 xv'' +
x^2v''' \\ \end{eqnarray*}
Substituting in (1) gives $$ x^5v''' +
9x^4 v'' + 19 x^3v'  =  0$$

Let $ w = v'$ then $$x^2w'' + 9xw' + 19 w = 0 \hspace{.7in} (2)$$

This is an Euler type equation.

Try $w = x^p.$ Substituting in (2) gives \begin{eqnarray*} (p(p-1)
+ 9p + 19) &amp; = &amp; 0 \\ \Rightarrow p^2 + 8p + 19 &amp; = &amp; 0 \\
\Rightarrow p &amp; = &amp; -4 \pm \sqrt 3 i \\ w &amp; = &amp; x^{-4 \pm \sqrt 3
i} \\ \Rightarrow v' &amp; =&amp; x^{-4 \pm \sqrt 3 i} \\ \Rightarrow v &amp;
= &amp; Cx^{-3 \pm \sqrt 3 i } \end{eqnarray*}
Now:

$\begin{array}{rl} x^{-3 + \sqrt 3 i} = x^{-3}x^{\sqrt 3 i} &amp; =
x^{-3}e^{\sqrt 3 i \ln x} \\ &amp; = x^{-3}(\cos (\sqrt 3 \ln x) + i
\sin (\sqrt 3 \ln x))\end{array}$

Hence the real $v$ is given by

$\ds v = x^{-3}\left(A\cos (\sqrt 3 \ln x) + B \sin (\sqrt 3 \ln
x)\right)$

and $y_2 = y_1v = x^2v = x^{-1}\left(A\cos (\sqrt 3 \ln x) + B\sin
(\sqrt 3 \ln x)\right)$

So the general solution is $$y = x^{-1}\left(A\cos (\sqrt 3 \ln x)
+ B \sin (\sqrt 3 \ln x)\right) + Cx^2$$


\end{description}



\end{document}
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