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)
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.