2.3.2.4 Example 4. \(x^{2}y^{\prime \prime }+3xy^{\prime }+4x^{4}y=0\)

Given

\begin{equation} x^{2}y^{\prime \prime }+3xy^{\prime }+4x^{4}y=0 \tag {1}\end{equation}

Expanding around \(x=0\). Writing the ode as

\[ y^{\prime \prime }+\frac {3}{x}y^{\prime }+4x^{2}y=0 \]

Shows that \(x=0\) is a singular point. But \(\lim _{x\rightarrow 0}x\frac {3}{x}=3\). 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{align} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+4x^{4}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+4} & =0 \tag {1A}\end{align}

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

\[ \sum _{n=0}^{\infty }4a_{n}x^{n+r+4}=\sum _{n=4}^{\infty }4a_{n-4}x^{n+r}\]

And now Eq (1A) becomes

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=4}^{\infty }4a_{n-4}x^{n+r}=0 \tag {1B}\end{equation}

\(n=0\) gives the indicial equation

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

Since \(a_{0}\neq 0\) then the above becomes

\[ \left ( r\right ) \left ( r-1\right ) +3r=0 \]

Hence the roots of the indicial equation are \(r_{1}=0,r_{2}=-2\). Or \(r_{1}=r_{2}+N\) where \(N=2\). We always take \(r_{1}\) to be the larger of the roots.

When this happens, the solution is given by

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

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

\begin{equation} y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+r} \tag {2}\end{equation}

Where we take \(a_{0}=1\) as it is arbitrary and where \(r=r_{1}=0\). 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}=Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r} \tag {3}\end{equation}

We will show below how to find \(C\) and \(b_{n}\). First, let us find \(y_{1}\left ( x\right ) \). From Eq(2)

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

We need to remember that in the above \(r\) is not a symbol any more. It will have the indicial root value, which is \(r=r_{1}=0\) in this case. But we keep \(r\) as symbol for now, in order to obtain \(a_{n}\left ( r\right ) \) as function of \(r\) first and use this to find \(b_{n}\left ( r\right ) \). At the very end we then evaluate everything at \(r=r_{1}=0\). Substituting the above in (1) gives Eq (1B) above (We are following pretty much the same process we did to find the indicial equation here)

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=4}^{\infty }4a_{n-4}x^{n+r}=0 \tag {1B}\end{equation}

Now we are ready to find \(a_{n}\). Now 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.

For \(n=1\), Eq (1B) gives

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

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

\[ 3a_{1}=0 \]

Hence \(a_{1}=0\).

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 \(b_{n}\).

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)

For \(n=2\), Eq (1B) gives

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

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

\begin{align*} \left ( \left ( 2\right ) \left ( 1\right ) +3\left ( 2\right ) \right ) a_{2} & =0\\ 8a_{2} & =0 \end{align*}

Hence \(a_{2}=0\). The table becomes

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=0\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2\)	\(0\)	\(0\)

For \(n=3\), Eq (1B) gives

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

But \(r=0\). The above becomes

\begin{align*} \left ( 3\right ) \left ( 2\right ) a_{3}+3\left ( 3\right ) a_{3} & =0\\ 15a_{3} & =0 \end{align*}

Hence \(a_{3}=0\) and the table becomes

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=0\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2\)	\(0\)	\(0\)
\(3\)	\(0\)	\(0\)

For \(n\geq 4\) we obtain the recursion equation

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+3\left ( n+r\right ) a_{n}+4a_{n-4} & =0\nonumber \\ \left ( \left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) \right ) a_{n}+4a_{n-4} & =0\nonumber \\ a_{n}\left ( r\right ) & =-\frac {4a_{n-4}\left ( r\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) } \tag {4}\end{align}

The above is very important, since we will use it to find \(b_{n}\left ( r\right ) \) later on. For now, we are just finding the \(a_{n}\). Now we find few more \(a_{n}\) terms. From (4) for \(n=4\)

\[ a_{4}\left ( r\right ) =-\frac {4a_{0}\left ( r\right ) }{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\]

and \(r=r_{1}=0\,\) and \(a_{0}=1\), then the above becomes

\[ a_{4}=-\frac {4}{\left ( 4\right ) \left ( 3\right ) +3\left ( 4\right ) }=-\frac {1}{6}\]

The table becomes

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

And for \(n=5\) from Eq(4)

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

Since \(a_{1}=0\). Similarly \(a_{6}=0,a_{7}=0\). For \(n=8\)

\[ a_{8}\left ( r\right ) =-\frac {4a_{4}\left ( r\right ) }{\left ( 8+r\right ) \left ( 8+r-1\right ) +3\left ( 8+r\right ) }\]

But \(a_{4}\left ( r\right ) =-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\). The above becomes

\[ a_{8}\left ( r\right ) =\frac {4\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }}{\left ( 8+r\right ) \left ( 8+r-1\right ) +3\left ( 8+r\right ) }=\frac {16}{r^{4}+28r^{3}+284r^{2}+1232r+1920}\]

When \(r=r_{1}=0\) the above becomes

\[ a_{8}\left ( r\right ) =\frac {1}{120}\]

And so on. The table becomes

\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=0\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2\)	\(0\)	\(0\)
\(3\)	\(0\)	\(0\)
\(4\)	\(-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\)	\(-\frac {1}{6}\)
\(5\)	\(0\)	\(0\)
\(6\)	\(0\)	\(0\)
\(7\)	\(0\)	\(0\)
\(8\)	\(\frac {16}{r^{4}+28r^{3}+284r^{2}+1232r+1920}\)	\(\frac {1}{120}\)

Hence \(y_{1}\left ( x\right ) \) is

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

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

\begin{align} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\tag {5}\\ & =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}+a_{7}x^{7}+a_{8}x^{8}+\cdots \nonumber \end{align}

Using values found for \(a_{n}\) in the above table, then (5) becomes

\begin{align*} y_{1} & =1+a_{4}x^{4}+a_{8}x^{8}+\cdots \\ & =1-\frac {1}{6}x^{4}+\frac {1}{120}x^{8}+O\left ( x^{9}\right ) \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

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

The first thing to do is to determine if \(C\) is zero or not. This is done by finding

\[ \lim _{r\rightarrow r_{2}}a_{N}\left ( r\right ) \]

If this limit exist, then \(C=0\), else we need to keep the log term. From the above above we see that \(a_{N}\left ( r\right ) =a_{2}\left ( r\right ) =0\). Recall that \(N=2\) since this was the difference between the two roots and \(r_{2}=-2\) (the smaller root). Therefore

\[ \lim _{r\rightarrow r_{2}}0=\lim _{r\rightarrow 0}0=0 \]

Hence the limit exist. Therefore we do not need the log term. This means we can let \(C=0\). This is the easy case. Hence (3) becomes

\begin{align} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r}\tag {3A}\\ & =x^{-2}\sum _{n=0}^{\infty }b_{n}x^{n}\nonumber \end{align}

Since \(r=r_{2}=-2\).  Let \(b_{0}=1\). We have to remember now that \(b_{N}=b_{2}=0\). This is the same we did when the log term was needed in the above example, since \(b_{N}\) is arbitrary, and used to generate \(y_{1}\left ( x\right ) \). Common practice is to use \(b_{N}=0\). The rest of the \(b_{n}\) are found in similar way, from recursive relation as was done above. Substituting (3A) into \(x^{2}y^{\prime \prime }+3xy^{\prime }+4x^{4}y=0\) gives Eq. (1B) again, but with \(a_{n}\) replaced by \(b_{n}\)

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) b_{n}x^{n+r}+\sum _{n=4}^{\infty }4b_{n-4}x^{n+r}=0 \tag {1B}\end{equation}

For \(n=0\,\), we skip and let \(b_{0}=1\). For \(n=1\) the above gives \(b_{1}=0\). And \(b_{2}=0\) since it is the special term \(b_{N}\). And for \(n=3\), we get \(b_{3}=0\). The table for \(b_{n}\) is now

\(n\)	\(b_{n}\left ( r\right ) \)	\(b_{n}\left ( r=-2\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2\)	\(0\)	\(0\)
\(3\)	\(0\)	\(0\)

For \(n\geq 4\), the recursion relation is

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) b_{n}+3\left ( n+r\right ) b_{n}+4b_{n-4} & =0\\ b_{n}\left ( r\right ) & =-\frac {4b_{n-4}\left ( r\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) }\end{align*}

For \(n=4\)

\begin{align*} b_{4}\left ( r\right ) & =-\frac {4b_{0}\left ( r\right ) }{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\\ & =-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\qquad b_{0}=1 \end{align*}

but \(r=-2\). The above becomes

\[ b_{4}=-\frac {4}{\left ( 4-2\right ) \left ( 4-2-1\right ) +3\left ( 4-2\right ) }=-\frac {1}{2}\]

The table becomes

\(n\)	\(b_{n}\left ( r\right ) \)	\(b_{n}\left ( r=-2\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2^{\ast }\)	\(0\)	\(0\)
\(3\)	\(0\)	\(0\)
\(4\)	\(-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\)	\(-\frac {1}{2}\)

We will find that \(b_{5}=b_{6}=b_{7}=0\). And for \(n=8\)

\[ b_{8}\left ( r\right ) =-\frac {4b_{4}\left ( r\right ) }{\left ( 8+r\right ) \left ( 7+r\right ) +3\left ( 8+r\right ) }\]

But \(b_{4}\left ( r\right ) =-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\). Hence

\[ b_{8}\left ( r\right ) =\frac {4\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }}{\left ( 8+r\right ) \left ( 7+r\right ) +3\left ( 8+r\right ) }=\frac {16}{r^{4}+28r^{3}+284r^{2}+1232r+1920}\]

But \(r=-2\).

\[ b_{8}\left ( r\right ) =\frac {16}{\left ( -2\right ) ^{4}+28\left ( -2\right ) ^{3}+284\left ( -2\right ) ^{2}+1232\left ( -2\right ) +1920}=\frac {1}{24}\]

The table becomes

\(n\)	\(b_{n}\left ( r\right ) \)	\(b_{n}\left ( r=-2\right ) \)
\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)
\(2^{\ast }\)	\(0\)	\(0\)
\(3\)	\(0\)	\(0\)
\(4\)	\(-\frac {4}{\left ( 4+r\right ) \left ( 4+r-1\right ) +3\left ( 4+r\right ) }\)	\(-\frac {1}{2}\)
\(5\)	\(0\)	\(0\)
\(6\)	\(0\)	\(0\)
\(7\)	\(0\)	\(0\)
\(8\)	\(\frac {16}{r^{4}+28r^{3}+284r^{2}+1232r+1920}\)	\(\frac {1}{24}\)

And so on. Hence the second solution is

\begin{align*} y_{2}\left ( x\right ) & =\sum _{n=0}^{\infty }b_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }b_{n}x^{n-2}\\ & =x^{-2}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =x^{-2}\left ( b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+b_{5}x^{5}+b_{6}x^{6}+b_{7}x^{7}+b_{8}x^{8}+\cdots \right ) \\ & =x^{-2}\left ( 1-\frac {1}{2}x^{4}+\frac {1}{24}b_{8}x^{8}+O\left ( x^{9}\right ) \right ) \end{align*}

Therefore the general solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {1}{6}x^{4}+\frac {1}{120}x^{8}+O\left ( x^{9}\right ) \right ) +c_{2}\left ( x^{-2}\left ( 1-\frac {1}{2}x^{4}+\frac {1}{24}b_{8}x^{8}+O\left ( x^{9}\right ) \right ) \right ) \end{align*}

The following are important items to remember. Always let \(b_{N}=0\) where \(N\) is the difference between the roots. When the log term is not needed (as in this problem), \(y_{2}\) is found in very similar way to \(y_{1}\) where \(b_{0}=1\) and the recursion formula is used to find all \(b_{n}\). But when the log term is needed (as in the above problem), it is a little more complicated and need to find \(C\) and \(b_{1}\) values by comparing coefficients as was done).

This completes the solution.