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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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{\bf Question}

Use the given solution to find a second solution for each of the
following differential equations:

\begin{tabular}{rlll}
a) &amp; $x^2y''+2xy'-2y=0$ &amp; $ y_1(x)=x$  &amp; (*)\\
%b) &amp; $ x^2y''+xy'+(x^2-1/4)y=0$ &amp; $y_1(x)=x^{-1/2}\sin x$\\
b) &amp; $(x-1)y''-xy'+y=0 \quad x&gt;1$ &amp; $y_1(x)=e^x$ &amp; (*)\\

\end{tabular}



\vspace{0.25in}

{\bf Answer}

Check that $y_1=x$ is a solution to the equation\\ This follows
since $y_1'=1$ and $y_1''=0$ and putting these into the equation
makes it true.\\ Equation is in standard form with $\ds
p(x)=\frac{2}{x}$ and $\ds q(x)=\frac{-2}{x^2}$. Use method of
reduction of order with $y_2=vy_1$ where $\ds v(x)=\int
\left(\frac{1}{y_1^2(x)}\;e^{-\int p(x) \;dx}\right)\;dx$\\ so
that $\ds v(x) =\int \left(\frac{1}{x^2}\;e^{-\int
\frac{2}{x}\;dx} \right)\;dx =\int \left(\frac{1}{x^2}\;e^{-2\ln
x} \right)\;dx =\int \left(\frac{1}{x^2}\;\frac{1}{x^2}\right)\;dx
=\frac{-1}{3x^3}$\\ Hence the general solution is $\ds y(x)= A x +
\frac{B}{x^2} $



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