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\begin{document}
{\Large

\begin{center}
{\bf Reduction of Order Method for Finding a Second Solution}
\end{center}

This lecture is based on the second half of Section 3.4. It will
concern linear second order equations with non-constant coefficients.
Suppose we want to find the general solution to
\[
y'' + p(t)y' + q(t)y = 0,\hspace{1in} (*)
\]
and somehow we know $y=f(t)$ is a solution. We need to find another
solution that is not a constant multiple of $f(t)$. Here is a trick.

Let $y(t) = v(t)f(t)$. We will plug this into $(*)$ and solve for $v$.
We compute the needed derivatives.
\begin{eqnarray*}
y &=& vf \\
y' &=& v'f + vf' \\
y'' &=& v''f + 2 v' f' + vf''
\end{eqnarray*}
Plugging into $(*)$ gives
\begin{eqnarray*}
v''f+2v'f' + vf'' + pv'f + pvf' + qvf &=& 0.
\end{eqnarray*}
Regrouping the terms gives
\begin{eqnarray*}
fv'' + (2f'+pf)v' + (f''+pf'+qf) &=& 0, \:\:\mbox{or}\\
fv'' + (2f'+pf)v' &=& 0 \:\:\:\mbox{(since $f$ is a solution).}
\end{eqnarray*}
Next let $w=v'$ and write
\[
f w' + (2f'+pf) w = 0. \hspace{1in} (@)
\]
Thus, we have reduced the order from two to one. Notice that the resulting equation is both linear and separable. Solve it for $w$, then
integrate $w$ to get $v$ and hope it is not a constant.

{\bf Example 1.}
Consider
\[
y'' - \frac{x}{x-1} y' + \frac{1}{x-1} y = 0. \hspace{1in} (\#)
\]
assume $x>1$. After staring at it for a while we notice $y_1(x)=x$ 
is a solution. Check this. Let $y = y_2(x) = v(x) x$. Then
$y=vx$, $y' = v'x + v$, and $y'' = v''x + 2v'$. Thus $(\#)$ becomes
\[
v''x + 2v' - \frac{x}{x-1}(v'x + v) + \frac{1}{x-1} vx = 0,
\]
or
\[
xv'' + \left(2 - \frac{x^2}{x-1}\right) v' + \left(\frac{-x}{x-1}+\frac{x}{x-1} \right)v = 0.
\]
Thus we have 
\[
v'' + \left(\frac{2}{x} - \frac{x}{x-1}\right) v' =0.
\]
Let $w=v'$. Then we have
\[
\frac{dw}{dx} + \left(\frac{2}{x} - \frac{x}{x-1}\right)w = 0.
\]
We will use that it is separable to get
\[
\int \frac{1}{w}\, dw = \int \left(\frac{x}{x-1} - \frac{2}{x} \right) \, dx.
\]
Thus, since $x/(x-1) = 1 + 1/(x-1)$, we have
\[
\ln | w |= x + \ln (x-1) - 2 \ln x + C.
\]
Therefore,
\[
w = Ce^x \left(\frac{x-1}{x^2}\right).
\]
Since we do not need the general solution for $v$ we let $C=1$.

We integrate $w$ to get $v$,
\begin{eqnarray*}
v &=& \int e^x \left(\frac{x-1}{x^2}\right) \, dx \\
&&\\
&=& \int \frac{e^x}{x}\, dx  - \int \frac{e^x}{x^2} \, dx 
\end{eqnarray*}
We use integration by parts on the second integral. Let $u = e^x$
and $dz = x^{-2} dx$. Then $du = e^x dx$ and $z = -x^{-1}$. Thus,
\[
\int \frac{e^x}{x^2} = -\frac{e^x}{x} + \int \frac{e^x}{x} \, dx
\]
Thus,
\[
v = \int \frac{e^x}{x}\, dx + \frac{e^x}{x} -\int \frac{e^x}{x} \, dx
=  \frac{e^x}{x}.
\]
Finally,
\[
y_2(x) =  \frac{e^x}{x} \cdot x = e^x,
\]
is our second solution. Plug it into $(\#)$ and check this!
The general solution is
\[
y(x) = C_1 x + C_2 e^x.
\]
So there!

One last point. Could it happen that we go through all this work and turns out that $v$ is a constant, so that we do not get a new solution that is not a multiple of the first? No. Because if $v$ is a 
constant then $w=0$. But we know that $(@)$ has nontrivial solutions
since otherwise solutions for many valid initial value problems 
would not exist.
\end{document}
