Solve\begin{equation} x^{2}y^{\prime \prime }-xy=0 \tag {1}\end{equation} Using power series method by expanding around \(x=0\). Writing the ode as\[ y^{\prime \prime }-\frac {1}{x}y=0 \] Shows that \(x=0\) is a singular point. But \(\lim _{x\rightarrow 0}x^{2}\frac {1}{x}=0\). 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{align} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-x\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 }a_{n}x^{n+r+1} & =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\[ \sum _{n=0}^{\infty }a_{n}x^{n+r+1}=\sum _{n=1}^{\infty }a_{n-1}x^{n+r}\] 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=1}^{\infty }a_{n-1}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}x^{r} & =0\\ \left ( r\right ) \left ( r-1\right ) a_{0}x^{r} & =0 \end{align*}
Since \(a_{0}\neq 0\) then the above becomes\[ \left ( r\right ) \left ( r-1\right ) x^{r}=0 \] Since this is true for all \(x\), then\[ \left ( r\right ) \left ( r-1\right ) =0 \] Hence the roots of the indicial equation are \(r_{1}=1,r_{2}=0\). Or \(r_{1}=r_{2}+N\) where \(N=1\). We always take \(r_{1}\) to be the larger of the roots.
When this happens, the solution is given by\[ y=c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \] 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 \(r=r_{1}=1\). 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}=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 \(r=r_{1}=1\) 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 \(r=r_{1}=1\). 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. We start from \(n=1\). For \(n\geq 1\) we obtain the recursion equation\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}-a_{n-1} & =0\\ a_{n} & =\frac {a_{n-1}}{\left ( n+r\right ) \left ( n+r-1\right ) }\end{align*}
To more clearly indicate that \(a_{n}\) is function of \(r\), we write the above as\begin{equation} a_{n}\left ( r\right ) =\frac {a_{n-1}\left ( r\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {4}\end{equation} 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 \(n=1\)\[ a_{1}\left ( r\right ) =\frac {a_{0}\left ( r\right ) }{\left ( 1+r\right ) \left ( r\right ) }=\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\qquad a_{0}=1 \] and \(r=r_{1}=1\) then the above becomes\[ a_{1}=\frac {1}{2}\] 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 \(b_{n}\).
| \(n\) | \(a_{n}\left ( r\right ) \) | \(a_{n}\left ( r=r_{1}\right ) \) |
| \(0\) | \(1\) | \(1\) |
| \(1\) | \(\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\) | \(\frac {1}{2}\) |
And for \(n=2\) from Eq(4)\[ a_{2}\left ( r\right ) =\frac {a_{1}\left ( r\right ) }{\left ( 2+r\right ) \left ( 1+r\right ) }\] But \(a_{1}\left ( r\right ) =\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\). Then \[ a_{2}\left ( r\right ) =\frac {\frac {1}{\left ( 1+r\right ) \left ( r\right ) }}{\left ( 2+r\right ) \left ( 1+r\right ) }=\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) }\] When \(r=r_{1}=1\) the above becomes\[ a_{2}=\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) }=\frac {1}{12}\] The table becomes
| \(n\) | \(a_{n}\left ( r\right ) \) | \(a_{n}\left ( r=r_{1}\right ) \) |
| \(0\) | \(1\) | \(1\) |
| \(1\) | \(\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\) | \(\frac {1}{2}\) |
| \(2\) | \(\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) }\) | \(\frac {1}{12}\) |
For \(n=3\) Eq (4) gives\[ a_{3}\left ( r\right ) =\frac {a_{2}\left ( r\right ) }{\left ( 3+r\right ) \left ( 2+r\right ) }\] Using the value of \(a_{2}\left ( r\right ) \) from the the above becomes\[ a_{3}\left ( r\right ) =\frac {\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) }}{\left ( 3+r\right ) \left ( 2+r\right ) }=\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) }\] When \(r=r_{1}=1\) the above becomes\[ a_{3}=\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}\left ( 4\right ) }=\frac {1}{144}\] The Table now becomes
| \(n\) | \(a_{n}\left ( r\right ) \) | \(a_{n}\left ( r=r_{1}\right ) \) |
| \(0\) | \(1\) | \(1\) |
| \(1\) | \(\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\) | \(\frac {1}{2}\) |
| \(2\) | \(\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) }\) | \(\frac {1}{12}\) |
| \(3\) | \(\frac {1}{r\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) }\) | \(\frac {1}{144}\) |
And so on. Hence \(y_{1}\left ( x\right ) \) is\[ y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+r}\] But \(r=r_{1}=1\). Therefore\begin{align} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+1}\tag {5}\\ & =x\sum _{n=0}^{\infty }a_{n}x^{n}\nonumber \\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \nonumber \\ & =x\left ( 1+\frac {1}{2}x+\frac {1}{12}x^{2}+\frac {1}{144}x^{3}+\cdots \right ) \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\begin{equation} y_{2}=Cy_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=0}^{\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 \(a_{N}=a_{1}=\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\). Recall that \(N=1\) since this was the difference between the two roots and \(r_{2}=0\) (the smaller root). Therefore\[ \lim _{r\rightarrow r_{2}}a_{1}\left ( r\right ) =\lim _{r\rightarrow 0}\frac {1}{\left ( 1+r\right ) \left ( r\right ) }\] Which does not exist. Therefore we need to keep the log term. In this case, we replace Eq. (3) back in the original ODE. \begin{align*} y_{2}^{\prime } & =Cy_{1}^{\prime }\ln \left ( x\right ) +Cy_{1}\frac {1}{x}+\sum _{n=0}^{\infty }\left ( n+r\right ) b_{n}x^{n+r-1}\\ y_{2}^{\prime \prime } & =Cy_{1}^{\prime \prime }\ln \left ( x\right ) +Cy_{1}^{\prime }\frac {1}{x}+Cy_{1}^{\prime }\frac {1}{x}-Cy_{1}\frac {1}{x^{2}}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r-2}\\ & =Cy_{1}^{\prime \prime }\ln \left ( x\right ) +2Cy_{1}^{\prime }\frac {1}{x}-Cy_{1}\frac {1}{x^{2}}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r-2}\end{align*}
Substituting the above in \(x^{2}y^{\prime \prime }-xy=0\) gives\begin{align*} x^{2}\left ( Cy_{1}^{\prime \prime }\ln \left ( x\right ) +2Cy_{1}^{\prime }\frac {1}{x}-Cy_{1}\frac {1}{x^{2}}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r-2}\right ) -x\left ( Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r}\right ) & =0\\ Cx^{2}y_{1}^{\prime \prime }\ln \left ( x\right ) +2x^{2}Cy_{1}^{\prime }\frac {1}{x}-Cy_{1}+x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r-2}-Cxy_{1}\ln \left ( x\right ) -x\sum _{n=0}^{\infty }b_{n}x^{n+r} & =0\\ Cx^{2}y_{1}^{\prime \prime }\ln \left ( x\right ) +2xCy_{1}^{\prime }-Cy_{1}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}-Cxy_{1}\ln \left ( x\right ) -\sum _{n=0}^{\infty }b_{n}x^{n+r+1} & =0\\ C\ln \left ( x\right ) \left ( x^{2}y_{1}^{\prime \prime }-xy_{1}\right ) +2xCy_{1}^{\prime }-Cy_{1}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}-\sum _{n=0}^{\infty }b_{n}x^{n+r+1} & =0 \end{align*}
But \(x^{2}y_{1}^{\prime \prime }-xy_{1}=0\) since \(y_{1}\) is solution to the ode. The above simplifies to\begin{equation} C\left ( 2xy_{1}^{\prime }-y_{1}\right ) +\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}-\sum _{n=0}^{\infty }b_{n}x^{n+r+1}=0 \tag {6}\end{equation} The above is what we will use to determine \(C\) and all the \(b_{n}\). Remembering that \(r=r_{2}=0\) in the above, since this is for the second solution associated with the second root which we found above to be zero. But we found \(y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+1}\) then\[ y_{1}^{\prime }=\sum _{n=0}^{\infty }\left ( n+1\right ) a_{n}x^{n}\] Eq (6) now becomes\begin{align*} C\left ( 2x\sum _{n=0}^{\infty }\left ( n+1\right ) a_{n}x^{n}\right ) -C\left ( \sum _{n=0}^{\infty }a_{n}x^{n+1}\right ) +\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}-\sum _{n=0}^{\infty }b_{n}x^{n+r+1} & =0\\ 2C\sum _{n=0}^{\infty }\left ( n+1\right ) a_{n}x^{n+1}-C\sum _{n=0}^{\infty }a_{n}x^{n+1}+\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) b_{n}x^{n+r}-\sum _{n=0}^{\infty }b_{n}x^{n+r+1} & =0 \end{align*}
But \(r=r_{2}=0\). The above becomes\[ 2C\sum _{n=0}^{\infty }\left ( n+1\right ) a_{n}x^{n+1}-C\sum _{n=0}^{\infty }a_{n}x^{n+1}+\sum _{n=0}^{\infty }n\left ( n-1\right ) b_{n}x^{n}-\sum _{n=0}^{\infty }b_{n}x^{n+1}=0 \] Adjusting the index of terms above, so so all \(x\) powers are the same gives\begin{equation} 2C\sum _{n=1}^{\infty }na_{n-1}x^{n}-C\sum _{n=1}^{\infty }a_{n-1}x^{n}+\sum _{n=0}^{\infty }n\left ( n-1\right ) b_{n}x^{n}-\sum _{n=1}^{\infty }b_{n-1}x^{n}=0 \tag {7}\end{equation} \(n=0\) is skipped, since \(b_{0}\) is arbitrary and can be taken as say \[ b_{0}=1 \] At \(n=1\), Eq(7) gives\[ 2Ca_{0}-Ca_{0}-b_{0}=0 \] But \(a_{0}=1,b_{0}=1\) hence the above becomes\[ C=1 \] For \(b_{N}=b_{1}\) we are free to select any value since it is arbitrary. The standard way is to choose\[ b_{1}=0 \] Now we find the rest of the \(b_{n}\) terms. From Eq(7), for \(n=2\), it gives\[ 2C\left ( 2a_{1}\right ) -Ca_{1}+2b_{2}-b_{1}=0 \] But \(C=1,b_{1}=0\) and \(a_{1}=\) \(\frac {1}{2}\) from table. Hence the above becomes\begin{align*} 2\left ( 2\frac {1}{2}\right ) -\frac {1}{2}+2b_{2} & =0\\ 2-\frac {1}{2}+2b_{2} & =0\\ b_{2} & =-\frac {3}{4}\end{align*}
And for \(n=3\) from Eq. (7) it gives\[ 2C\left ( 3a_{2}\right ) -Ca_{2}+\left ( 3\right ) \left ( 2\right ) b_{3}-b_{2}=0 \] But \(C=1,b_{2}=-\frac {3}{4},a_{2}=\frac {1}{12}\). The above becomes\begin{align*} 2\left ( 3\left ( \frac {1}{12}\right ) \right ) -\frac {1}{12}+\left ( 3\right ) \left ( 2\right ) b_{3}+\frac {3}{4} & =0\\ b_{3} & =-\frac {7}{36}\end{align*}
And so on. Hence the second solution is, for \(r=0,C=1\)\begin{align*} y_{2}\left ( x\right ) & =Cy_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r}\\ & =y_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =y_{1}\left ( x\right ) \ln \left ( x\right ) +\left ( b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+\cdots \right ) \\ & =y_{1}\left ( x\right ) \ln \left ( x\right ) +\left ( 1+\left ( 0\right ) x-\frac {3}{4}x^{2}-\frac {7}{36}x^{3}+\cdots \right ) \\ & =\left ( x+\frac {1}{2}x^{2}+\frac {1}{12}x^{3}+\frac {1}{144}x^{4}+\cdots \right ) \ln x+\left ( 1+\frac {3}{4}x^{2}-\frac {7}{36}x^{3}+\cdots \right ) \\ & =x\left ( 1+\frac {1}{2}x+\frac {1}{12}x^{2}+\frac {1}{144}x^{3}+O\left ( x^{4}\right ) \right ) \ln x+\left ( 1+\frac {3}{4}x^{2}-\frac {7}{36}x^{3}+O\left ( x^{4}\right ) \right ) \end{align*}
Some observations: \(b_{N}\) is always taken as zero. Where \(N\) is the difference between the roots. In this case it is \(b_{1}=0\). Now that we found \(y_{1},y_{2}\) then the general solution is
This completes the solution.