2.2.2.3 Example 3. \(\left ( e^{x}-1\right ) y^{\prime \prime }+e^{x}y^{\prime }+y=0\)

Solve

\begin{equation} \left ( e^{x}-1\right ) y^{\prime \prime }+e^{x}y^{\prime }+y=0 \tag {1}\end{equation}

Using power series method by expanding around \(x=0\). Writing the ode as

\[ y^{\prime \prime }+\frac {e^{x}}{\left ( e^{x}-1\right ) }y^{\prime }+\frac {1}{\left ( e^{x}-1\right ) }y=0 \]

Shows that \(x=0\) is a singular point. But \(\lim _{x\rightarrow 0}x\frac {e^{x}}{\left ( e^{x}-1\right ) }=1\) and \(\lim _{x\rightarrow 0}x^{2}\frac {1}{\left ( e^{x}-1\right ) }=0\) Hence the singularity is removable. This means \(x=0\) is a regular singular point. In this case the Frobenius power series will be used instead of the standard power series. Let

\[ y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\]

Where \(r\) is to be determined. It is the root of the indicial equation. Therefore

\begin{align*} y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

Substituting the above in (1) gives

\begin{equation} \left ( e^{x}-1\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+e^{x}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0\nonumber \end{equation}

Expanding \(e^{x}\) in Taylor series around \(x\) gives \(e^{x}=1+x+\frac {1}{2}x^{2}+\frac {1}{6}x^{3}+\frac {1}{24}x^{4}+\cdots \). The above becomes

\begin{multline} \left ( x+\frac {1}{2}x^{2}+\frac {1}{6}x^{3}+\frac {1}{24}x^{4}+\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\nonumber \\ +\left ( 1+x+\frac {1}{2}x^{2}+\frac {1}{6}x^{3}+\frac {1}{24}x^{4}+\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \tag {1}\end{multline}

Expanding gives (and keeping only terns up to \(x^{4}\) gives

\begin{multline*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\frac {1}{2}x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\frac {1}{6}x^{3}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\\ +\frac {1}{24}x^{4}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\frac {1}{2}x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ +\frac {1}{6}x^{3}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\frac {1}{24}x^{4}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

Moving the \(x\) inside the sum, the above becomes

\begin{multline*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\frac {1}{2}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+1}\\ +\sum _{n=0}^{\infty }\frac {1}{24}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\frac {1}{2}\left ( n+r\right ) a_{n}x^{n+r+1}\\ +\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) a_{n}x^{n+r+2}+\sum _{n=0}^{\infty }\frac {1}{24}\left ( n+r\right ) a_{n}x^{n+r+3}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

Here, we need to make all powers on \(x\) the same, without making the sums start below zero. This can be done by adjusting the last term above as follows

\begin{multline} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }\frac {1}{2}\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}x^{n+r-1}+\sum _{n=2}^{\infty }\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}x^{n+r-1}\tag {2}\\ +\sum _{n=3}^{\infty }\frac {1}{24}\left ( n+r-3\right ) \left ( n+r-4\right ) a_{n-3}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r-1}+\sum _{n=2}^{\infty }\frac {1}{2}\left ( n+r-2\right ) a_{n-2}x^{n+r-1}\nonumber \\ +\sum _{n=3}^{\infty }\frac {1}{6}\left ( n+r-3\right ) a_{n-3}x^{n+r-1}+\sum _{n=4}^{\infty }\frac {1}{24}\left ( n+r-4\right ) a_{n-4}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0\nonumber \end{multline}

The case \(n=0\) gives the indicial equation

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) & =0\\ \left ( r\right ) \left ( r-1\right ) +\left ( r\right ) & =0\\ r^{2} & =0 \end{align*}

Hence the roots of the indicial equation are \(r=0\) which is a double root. Hence \(r_{1}=r_{2}=0\). When this happens, the solution is given by

\[ y=c_{1}y_{1}+c_{2}y_{2}\]

Where \(y_{1}\) is the first solution, which is assumed to be

\begin{equation} y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+r} \tag {3}\end{equation}

Where we take \(a_{0}=1\) as it is arbitrary and where \(r=r_{1}\). This is the standard Frobenius power series, just like we did to find the indicial equation, the only difference is that now we use \(r=r_{1}\), and hence it is a known value. Once we find \(y_{1}\left ( x\right ) \), then the second solution is

\begin{equation} y_{2}=y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r} \tag {3}\end{equation}

Something important to notice. In the sum above, it starts from \(1\) and not from \(0\). The main issue is how to find \(b_{n}\). Since that is the only thing we need to be able to complete the solution as \(y_{1}\left ( x\right ) \) is easily found.  It turns out that there is a relation between the \(b_{n}\) and the \(a_{n}\). The \(b_{n}\) can be found by taking just derivative of \(a_{n}\) as function of \(r\) for each \(n\) and then evaluate the result at \(r=r_{1}\). How this is done will be shown below. First we need to find \(y_{1}\left ( x\right ) \). We take Eq(3) and substitute it in the original ODE. This will result in Eq (2) which we found above so no need to repeat that. We just need to remember that now we now what \(r\) is. It has a numerical value unlike the above phase where we still did not know its value.

Now we are ready to find \(a_{n}\). We skip \(n=0\) since that was used to obtain the indicial equation, and we know that \(a_{0}=1\) is an arbitrary value to choose. We start from \(n=1\).

For \(n=1\) only, using Eq (2) gives

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) a_{1}+\frac {1}{2}\left ( n+r-1\right ) \left ( n+r-2\right ) a_{0}+\left ( n+r\right ) a_{1}+\left ( n+r-1\right ) a_{0}+a_{0} & =0\\ \left ( 1+r\right ) \left ( 1+r-1\right ) a_{1}+\frac {1}{2}\left ( 1+r-1\right ) \left ( 1+r-2\right ) a_{0}+\left ( 1+r\right ) a_{1}+\left ( 1+r-1\right ) a_{0}+a_{0} & =0\\ \left ( \left ( 1+r\right ) \left ( 1+r-1\right ) +\left ( 1+r\right ) \right ) a_{1}+\left ( \frac {1}{2}\left ( 1+r-1\right ) \left ( 1+r-2\right ) +\left ( 1+r-1\right ) +1\right ) a_{0} & =0 \end{align*}

But \(a_{0}=1\). The above becomes

\begin{align*} \left ( \left ( 1+r\right ) \left ( 1+r-1\right ) +\left ( 1+r\right ) \right ) a_{1} & =-\left ( \frac {1}{2}\left ( 1+r-1\right ) \left ( 1+r-2\right ) +\left ( 1+r-1\right ) +1\right ) \\ a_{1} & =\frac {-\left ( \frac {1}{2}\left ( 1+r-1\right ) \left ( 1+r-2\right ) +\left ( 1+r-1\right ) +1\right ) }{\left ( \left ( 1+r\right ) \left ( 1+r-1\right ) +\left ( 1+r\right ) \right ) }\\ & =-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\end{align*}

Which at \(r=0\) gives

\[ a_{1}=-1 \]

It is a good idea to use a table to keep record of the \(a_{n}\) values as function of \(r\), since this will be used later to find \(b_{n}\).

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\)	\(-1\)

For \(n=2\) only, using Eq (2) gives

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) a_{2}+\frac {1}{2}\left ( n+r-1\right ) \left ( n+r-2\right ) a_{1}+\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{0}+\left ( n+r\right ) a_{2}+\left ( n+r-1\right ) a_{1}+\frac {1}{2}\left ( n+r-2\right ) a_{0}+a_{1} & =0\\ \left ( 2+r\right ) \left ( 2+r-1\right ) a_{2}+\frac {1}{2}\left ( 2+r-1\right ) \left ( 2+r-2\right ) a_{1}+\frac {1}{6}\left ( 2+r-2\right ) \left ( 2+r-3\right ) a_{0}+\left ( 2+r\right ) a_{2}+\left ( 2+r-1\right ) a_{1}+\frac {1}{2}\left ( 2+r-2\right ) a_{0}+a_{1} & =0\\ \left ( \left ( 2+r\right ) \left ( 2+r-1\right ) +\left ( 2+r\right ) \right ) a_{2}+\left ( \frac {1}{2}\left ( 2+r-1\right ) \left ( 2+r-2\right ) +\left ( 2+r-1\right ) +1\right ) a_{1}+\left ( \frac {1}{6}\left ( 2+r-2\right ) \left ( 2+r-3\right ) +\frac {1}{2}\left ( 2+r-2\right ) \right ) a_{0} & =0\\ \left ( r+2\right ) ^{2}a_{2}+\left ( \frac {1}{2}r^{2}+\frac {3}{2}r+2\right ) a_{1}+\frac {1}{6}r\left ( r+2\right ) a_{0} & =0 \end{align*}

But \(a_{0}=1\) and \(a_{1}=-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\). The above becomes

\begin{align*} \left ( r+2\right ) ^{2}a_{2}+\left ( \frac {1}{2}r^{2}+\frac {3}{2}r+2\right ) \left ( -\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\right ) +\frac {1}{6}r\left ( r+2\right ) & =0\\ \left ( r+2\right ) ^{2}a_{2} & =\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{2\left ( r+1\right ) ^{2}}\\ a_{2} & =\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\end{align*}

At \(r=0\) the above becomes

\[ a_{2}=\frac {24}{12\left ( 2\right ) ^{2}}=\frac {1}{2}\]

The table becomes

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\)	\(-1\)
\(2\)	\(\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\)	\(\frac {1}{2}\)

For \(n=3\) only, using Eq (2) gives

\begin{multline*} \left ( n+r\right ) \left ( n+r-1\right ) a_{3}+\frac {1}{2}\left ( n+r-1\right ) \left ( n+r-2\right ) a_{2}+\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{1}\\ +\frac {1}{24}\left ( n+r-3\right ) \left ( n+r-4\right ) a_{0}+\left ( n+r\right ) a_{3}+\left ( n+r-1\right ) a_{2}+\frac {1}{2}\left ( n+r-2\right ) a_{1}+\frac {1}{6}\left ( n+r-3\right ) a_{0}+a_{2}=0 \end{multline*}

Or

\[ \left ( 3+r\right ) \left ( 2+r\right ) a_{3}+\frac {1}{2}\left ( 2+r\right ) \left ( 1+r\right ) a_{2}+\left ( 3+r\right ) a_{3}+\left ( 2+r\right ) a_{2}+\frac {1}{2}\left ( 1+r\right ) a_{1}+a_{2}=0 \]

Or

\begin{align*} \left ( \left ( 3+r\right ) \left ( 2+r\right ) +\left ( 3+r\right ) \right ) a_{3}+\left ( \frac {1}{2}\left ( 2+r\right ) \left ( 1+r\right ) +\left ( 2+r\right ) +1\right ) a_{2}+\frac {1}{2}\left ( 1+r\right ) a_{1} & =0\\ \left ( r+3\right ) ^{2}a_{3}+\left ( \frac {1}{2}r^{2}+\frac {5}{2}r+4\right ) a_{2}+\frac {1}{2}\left ( 1+r\right ) a_{1} & =0 \end{align*}

But \(a_{1}=-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2},a_{2}=\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\). The above becomes

\begin{align*} \left ( r+3\right ) ^{2}a_{3} & =-\left ( \frac {1}{2}r^{2}+\frac {5}{2}r+4\right ) a_{2}-\frac {1}{2}\left ( 1+r\right ) a_{1}\\ & =-\left ( \frac {1}{2}r^{2}+\frac {5}{2}r+4\right ) \left ( \frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\right ) -\frac {1}{2}\left ( 1+r\right ) \left ( -\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\right ) \\ & =-\frac {\left ( r^{6}+3r^{5}+9r^{4}+53r^{3}+158r^{2}+208r+144\right ) }{24\left ( r^{2}+3r+2\right ) ^{2}}\\ a_{3} & =-\frac {\left ( r^{6}+3r^{5}+9r^{4}+53r^{3}+158r^{2}+208r+144\right ) }{24\left ( r^{2}+3r+2\right ) ^{2}\left ( r+3\right ) ^{2}}\end{align*}

For \(r=0\) the above reduces to

\[ a_{3}=-\frac {144}{24\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}}=-\frac {1}{6}\]

The table becomes

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\)	\(-1\)
\(2\)	\(\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\)	\(\frac {1}{2}\)
\(3\)	\(-\frac {\left ( r^{6}+3r^{5}+9r^{4}+53r^{3}+158r^{2}+208r+144\right ) }{24\left ( r^{2}+3r+2\right ) ^{2}\left ( r+3\right ) ^{2}}\)	\(-\frac {1}{6}\)

And so on.  Recursion starts at \(n\geq 5\) but we have enough terms, so we stop here. \(y_{1}\left ( x\right ) \) is

\[ y_{1}\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n+r}\]

But \(r=r_{1}=0\). Therefore

\begin{align} y_{1}\left ( x\right ) & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+a_{6}x^{6}+\cdots \tag {6A}\\ & =1-x+\frac {1}{2}x^{3}-\frac {1}{6}x^{3}+\cdots \nonumber \end{align}

We are done finding \(y_{1}\left ( x\right ) \). This was not bad at all. Now comes the hard part. Which is finding \(y_{2}\left ( x\right ) \). From (3) it is given by

\[ y_{2}\left ( x\right ) =y_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r}\]

To find \(b_{n}\), we will use the following

\begin{equation} b_{n}\left ( r\right ) =\frac {d}{dr}\left ( a_{n}\left ( r\right ) \right ) \tag {7}\end{equation}

Notice that \(n\) starts from \(1\). Hence

\[ b_{1}\left ( r\right ) =\left . \frac {d}{dr}\left ( a_{1}\left ( r\right ) \right ) \right \vert _{r=r_{1}}\]

What the above says, is that we first take derivative of \(a_{n}\left ( r\right ) \) w.r.t. \(r\) and evaluate the result at the root of the indicial equation. Using the table above, we obtain (recalling that \(r_{1}=0\) in this example)

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}=0\right ) \)	\(b_{n}\left ( r\right ) =\frac {d}{dr}\left ( a_{n}\left ( r\right ) \right ) \)	\(b_{n}\left ( r=r_{1}=0\right ) \)
\(0\)	\(1\)	\(1\)	N/A since \(b\) starts from \(n=1\)	N/A
\(1\)	\(-\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\)	\(-1\)	\(\frac {d}{dr}\left ( -\frac {\left ( r^{2}+r+2\right ) }{2r^{2}+4r+2}\right ) =-\frac {\left ( r-3\right ) }{2\left ( r+1\right ) ^{3}}\)	\(\frac {3}{2}\)
\(2\)	\(\frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\)	\(\frac {1}{2}\)	\(\frac {d}{dr}\left ( \frac {r^{4}+4r^{3}+17r^{2}+26r+24}{12\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\right ) =\) \(-\frac {\left ( -r^{4}+7r^{3}+27r^{2}+53r+46\right ) }{6\left ( r^{2}+3r+2\right ) ^{3}}\)	\(-\frac {\left ( 46\right ) }{6\left ( 2\right ) ^{3}}=-\frac {23}{24}\)
\(3\)	\(-\frac {\left ( r^{6}+3r^{5}+9r^{4}+53r^{3}+158r^{2}+208r+144\right ) }{24\left ( r^{2}+3r+2\right ) ^{2}\left ( r+3\right ) ^{2}}\)	\(-\frac {1}{6}\)	\(\frac {d}{dr}\left ( -\frac {\left ( r^{6}+3r^{5}+9r^{4}+53r^{3}+158r^{2}+208r+144\right ) }{24\left ( r^{2}+3r+2\right ) ^{2}\left ( r+3\right ) ^{2}}\right ) \allowbreak =\frac {\left ( -9r^{7}-44r^{6}+24r^{5}+662r^{4}+2137r^{3}+3654r^{2}+3848r+1920\right ) }{24\left ( r^{3}+6r^{2}+11r+6\right ) ^{3}}\allowbreak \)	\(\frac {10}{27}\)

We have found all \(b_{n}\) terms. Hence

\[ y_{2}=y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r}\]

And since \(r=r_{1}=0\) then

\[ y_{2}=y_{1}\ln \left ( x\right ) +\left ( b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+b_{5}x^{5}+b_{6}x^{6}+\cdots \right ) \]

But from the above table, we see that \(b_{1}=\frac {3}{2},b_{2}=-\frac {23}{24},b_{3}=\frac {3}{8},\) The above becomes

\[ y_{2}=y_{1}\ln \left ( x\right ) +\left ( \frac {3}{2}x-\frac {23}{24}x^{2}+\frac {10}{27}x^{3}+O\left ( x^{4}\right ) \right ) \]

And we know what \(y_{1}\left ( x\right ) \) is from Eq (6A). Hence

\[ y_{2}=\left ( 1-x+\frac {1}{2}x^{3}-\frac {1}{6}x^{3}+O\left ( x^{4}\right ) \right ) \ln \left ( x\right ) +\left ( \frac {3}{2}x-\frac {23}{24}x^{2}+\frac {10}{27}x^{3}+O\left ( x^{4}\right ) \right ) \]

Therefore the general solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-x+\frac {1}{2}x^{3}-\frac {1}{6}x^{3}+O\left ( x^{4}\right ) \right ) +c_{2}\left ( \left ( 1-x+\frac {1}{2}x^{3}-\frac {1}{6}x^{3}+O\left ( x^{4}\right ) \right ) \ln \left ( x\right ) +\left ( \frac {3}{2}x-\frac {23}{24}x^{2}+\frac {10}{27}x^{3}+O\left ( x^{4}\right ) \right ) \right ) \end{align*}

This completes the solution.