2.4.1.4 Example 4. \(x^{2}y^{\prime \prime }-xy=0\)
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)
\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}
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
\begin{align*} y & =C_{1}y_{1}+C_{2}y_{2}\\ & =C_{1}x\left ( 1+\frac {1}{2}x+\frac {1}{12}x^{2}+\frac {1}{144}x^{3}+O\left ( x^{4}\right ) \right ) +C_{2}\left ( 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 ) \right ) \end{align*}
This completes the solution.