2.4.1.1 Example 1 \(x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=0\)

\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=0 \]

Comparing the ode to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {x^{2}-4}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}-4=-4\). Hence the indicial equation is

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

Therefore \(r_{1}=2,r_{2}=-2\).   Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ 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*}

The ode becomes

\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 }\left ( n+r\right ) a_{n}x^{n+r-1}+\left ( x^{2}-4\right ) \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 }\left ( n+r\right ) a_{n}x^{n+r}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-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 }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }4a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0\nonumber \end{align}

Re indexing to lowest powers on \(x\) gives

\begin{equation} \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r}=0 \tag {2}\end{equation}

\(n=0\) gives

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

Since \(a_{0}\neq 0\), then \(r^{2}=4\) or \(r_{1}=2,r_{2}=-2\) as was found above.  Since roots differ by an integer \(N=4\) then the two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

Where \(C\) can be zero depending on if \(\lim _{r\rightarrow r_{2}}a_{N}\left ( r\right ) \) exist or not. We start by finding \(y_{1}\). Substituting \(y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+r}\) into the ode gives (2) as was done above but with \(r=r_{1}.\) We start by \(n=1\) since \(n=0\) was used to find the roots.

For \(n=1\), EQ. (2) gives

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

Since \(r=2\) now, then

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

The recurrence relation is when \(n\geq 2\) from (2) is given by

\begin{align} \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) } \tag {4}\end{align}

Where in the above \(r=r_{1}=2\). For \(n=2\) the above gives

\begin{align*} a_{2} & =\frac {-a_{0}}{\left ( \left ( 2+r\right ) \left ( 1+r\right ) +\left ( 2+r\right ) -4\right ) }\\ & =-\frac {1}{r}\frac {a_{0}}{r+4}\\ & =-\frac {1}{2}\frac {a_{0}}{2+4}\\ & =-\frac {1}{12}a_{0}\end{align*}

For \(n=3\)

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

Since \(a_{1}=0\). For \(n=4\)

\begin{align*} a_{4} & =\frac {-a_{2}}{\left ( \left ( 4+r\right ) \left ( 3+r\right ) +\left ( 4+r\right ) -4\right ) }\\ & =-\frac {a_{2}}{\left ( r+6\right ) \left ( r+2\right ) }\\ & =-\frac {-\frac {1}{r}\frac {a_{0}}{r+4}}{\left ( r+6\right ) \left ( r+2\right ) }\\ & =\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }\\ & =\frac {1}{2}\frac {a_{0}}{\left ( 2+4\right ) \left ( 2+6\right ) \left ( 2+2\right ) }\\ & =\frac {1}{384}a_{0}\end{align*}

And so on. Hence

\begin{align*} y_{1} & =x^{r_{1}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =x^{2}\left ( a_{0}-\frac {1}{12}a_{0}x^{2}+\frac {1}{384}a_{0}x^{4}+\cdots \right ) \\ & =a_{0}x^{2}\left ( 1-\frac {1}{12}x^{2}+\frac {1}{384}x^{4}+\cdots \right ) \end{align*}

Or for \(a_{0}=1\)

\[ y_{1}=x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \]

Now we find \(y_{2}\). First we check if \(C=0\) or not. Since \(N=4\) then from the above we see

\begin{align*} a_{N}\left ( r\right ) & =a_{4}\left ( r\right ) \\ & =\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }\end{align*}

Hence

\begin{align*} \lim _{r\rightarrow r_{2}}a_{4} & =\lim _{r\rightarrow -2}a_{4}\\ & =\frac {1}{-2}\frac {a_{0}}{\left ( -2+4\right ) \left ( -2+6\right ) \left ( -2+2\right ) }\end{align*}

Which is not defined. Therefore \(C\) is not zero and we need the log term. The value of \(C\) is found when evaluating \(b_{N}\) below. Therefore we have

\begin{align*} y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n-2}\end{align*}

Hence

\begin{align*} y_{2}^{\prime } & =C\left ( y_{1}^{\prime }\ln \left ( x\right ) +y_{1}\frac {1}{x}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-3}\\ y_{2}^{\prime \prime } & =C\left ( y_{1}^{\prime \prime }\ln \left ( x\right ) +y_{1}^{\prime }\frac {1}{x}+y_{1}^{\prime }\frac {1}{x}-y_{1}\frac {1}{x^{2}}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-4}\\ & =C\left ( y_{1}^{\prime \prime }\ln \left ( x\right ) +2y_{1}^{\prime }\frac {1}{x}-y_{1}\frac {1}{x^{2}}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-4}\end{align*}

Substituting the above in \(x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=0\) gives

\begin{align*} x^{2}\left ( C\left ( y_{1}^{\prime \prime }\ln \left ( x\right ) +2y_{1}^{\prime }\frac {1}{x}-y_{1}\frac {1}{x^{2}}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-4}\right ) +x\left ( C\left ( y_{1}^{\prime }\ln \left ( x\right ) +y_{1}\frac {1}{x}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-3}\right ) +\left ( x^{2}-4\right ) \left ( Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n-2}\right ) & =0\\ x^{2}C\left ( y_{1}^{\prime \prime }\ln \left ( x\right ) +2y_{1}^{\prime }\frac {1}{x}-y_{1}\frac {1}{x^{2}}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+xC\left ( y_{1}^{\prime }\ln \left ( x\right ) +y_{1}\frac {1}{x}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\left ( x^{2}-4\right ) Cy_{1}\ln \left ( x\right ) +\left ( x^{2}-4\right ) \sum _{n=0}^{\infty }b_{n}x^{n-2} & =0\\ C\left ( x^{2}y_{1}^{\prime \prime }\ln \left ( x\right ) +2xy_{1}^{\prime }-y_{1}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+C\left ( xy_{1}^{\prime }\ln \left ( x\right ) +y_{1}\right ) +\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+x^{2}Cy_{1}\ln x-4Cy_{1}\ln \left ( x\right ) +x^{2}\sum _{n=0}^{\infty }b_{n}x^{n-2}-4\sum _{n=0}^{\infty }b_{n}x^{n-2} & =0\\ Cx^{2}y_{1}^{\prime \prime }\ln \left ( x\right ) +2Cxy_{1}^{\prime }-Cy_{1}+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+xCy_{1}^{\prime }\ln \left ( x\right ) +Cy_{1}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+x^{2}Cy_{1}\ln x-4Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n}-\sum _{n=0}^{\infty }4b_{n}x^{n-2} & =0\\ C\ln \left ( x\right ) \left [ x^{2}y_{1}^{\prime \prime }+xy_{1}^{\prime }+\left ( x^{2}-4\right ) y_{1}\right ] +2Cxy_{1}^{\prime }+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }b_{n}x^{n}-\sum _{n=0}^{\infty }4b_{n}x^{n-2} & =0 \end{align*}

But \(x^{2}y_{1}^{\prime \prime }+xy_{1}^{\prime }+\left ( x^{2}-4\right ) y_{1}=0\) since \(y_{1}\) is solution to the ode. The above simplifies to

\begin{equation} 2Cxy_{1}^{\prime }+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }b_{n}x^{n}-\sum _{n=0}^{\infty }4b_{n}x^{n-2}=0 \tag {5}\end{equation}

The above is what we will use to determine \(C\) and all the \(b_{n}\). But \(y_{1}=\sum _{n=0}^{\infty }a_{n}x^{n+2}\) then

\[ y_{1}^{\prime }=\sum _{n=0}^{\infty }\left ( n+2\right ) a_{n}x^{n+1}\]

Eq (5) now becomes

\begin{align*} 2Cx\sum _{n=0}^{\infty }\left ( n+2\right ) a_{n}x^{n+1}+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }b_{n}x^{n}-\sum _{n=0}^{\infty }4b_{n}x^{n-2} & =0\\ C\sum _{n=0}^{\infty }2\left ( n+2\right ) a_{n}x^{n+2}+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }b_{n}x^{n}-\sum _{n=0}^{\infty }4b_{n}x^{n-2} & =0 \end{align*}

Adjusting all the powers of \(x\) to the lowest one gives

\begin{equation} C\sum _{n=4}^{\infty }2\left ( n-2\right ) a_{n-4}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) \left ( n-3\right ) b_{n}x^{n-2}+\sum _{n=0}^{\infty }\left ( n-2\right ) b_{n}x^{n-2}+\sum _{n=2}^{\infty }b_{n-2}x^{n-2}-\sum _{n=0}^{\infty }4b_{n}x^{n-2}=0 \tag {6}\end{equation}

For \(n=0\)

\begin{align*} \left ( n-2\right ) \left ( n-3\right ) b_{0}+\left ( n-2\right ) b_{0}-4b_{0} & =0\\ \left ( -2\right ) \left ( -3\right ) b_{0}+\left ( -2\right ) b_{0}-4b_{0} & =0\\ 0b_{0} & =0 \end{align*}

Hence \(b_{0}\) is arbitrary. We always take \(b_{0}=1\). For \(n=1\)

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

For \(n=2\), EQ. (6) gives

\begin{align*} b_{0}-4b_{2} & =0\\ b_{2} & =\frac {b_{0}}{4}\\ & =\frac {1}{4}\end{align*}

For \(n=3\)

\begin{align*} b_{3}+b_{1}-4b_{3} & =0\\ -3b_{3} & =0\\ b_{3} & =0 \end{align*}

Now we get to \(n=4\) which is the special case since \(N=4\). This will generate the value of \(C\). From (6) and for \(n=4\)

\begin{align*} 4Ca_{0}+2b_{4}+2b_{4}+b_{2}-4b_{4} & =0\\ 4Ca_{0} & =-b_{2}\\ C & =-\frac {b_{2}}{4a_{0}}\\ & =-\frac {\frac {1}{4}}{4}\\ & =-\frac {1}{16}\end{align*}

We also notice that \(b_{4}\) is not used as it cancels. We always set \(b_{N}=b_{4}=0\) in this case.  Now that we found \(C\) we can use recursive relation to all higher values of \(b_{n}\). Form (6) and for all \(n>4\) we have

\begin{align} 2C\left ( n-2\right ) a_{n-4}+\left ( n-2\right ) \left ( n-3\right ) b_{n}+\left ( n-2\right ) b_{n}+b_{n-2}-4b_{n} & =0\nonumber \\ b_{n} & =\frac {-2C\left ( n-2\right ) a_{n-4}-b_{n-2}}{\left ( n-2\right ) \left ( n-3\right ) +\left ( n-2\right ) -4} \tag {7}\end{align}

In the above, we already know all the \(a_{n}\) values since we solved for \(y_{1}\) before. We also know \(C\). For \(n=5\) EQ. (7) gives

\begin{align*} b_{5} & =\frac {-2C\left ( 3\right ) a_{1}-b_{3}}{\left ( 3\right ) \left ( 2\right ) +\left ( 3\right ) -4}\\ & =\frac {-6Ca_{1}-b_{3}}{5}\end{align*}

But \(b_{3}=0\) and \(a_{1}=0\) hence \(b_{5}=0\). For \(n=6\) EQ. (7) gives

\begin{align*} b_{6} & =\frac {-2C\left ( 4\right ) a_{2}-b_{4}}{\left ( 4\right ) \left ( 3\right ) +\left ( 4\right ) -4}\\ & =\frac {-8Ca_{2}-b_{4}}{12}\end{align*}

but \(b_{4}=0,a_{2}=-\frac {1}{12}a_{0}=-\frac {1}{12}\) since \(a_{0}=1\) and \(C=\frac {1}{16}\). The above gives

\begin{align*} b_{6} & =\frac {-8\left ( -\frac {1}{16}\right ) \left ( -\frac {1}{12}\right ) }{12}\\ & =-\frac {1}{288}\end{align*}

And so on. Hence

\begin{align*} y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n-2}\\ & =-\frac {1}{12}y_{1}\ln \left ( x\right ) +x^{-2}\left ( b_{0}+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+b_{5}x^{5}+b_{6}x^{6}+\cdots \right ) \\ & =-\frac {1}{12}y_{1}\ln \left ( x\right ) +x^{-2}\left ( 1+\frac {1}{4}x^{2}-\frac {1}{288}x^{6}+\cdots \right ) \\ & =-\frac {1}{12}y_{1}\ln \left ( x\right ) +\left ( \frac {1}{x^{2}}+\frac {1}{4}-\frac {1}{288}x^{4}+\cdots \right ) \end{align*}

Therefore the final solution is

\begin{align} y & =c_{1}y_{1}+c_{2}y_{2}\nonumber \\ & =c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +x^{-2}\left ( 1+\frac {1}{4}x^{2}-\frac {1}{288}x^{6}+\cdots \right ) \right ) \tag {8}\end{align}

Let us now try different IC on the above.

IC \(y\left ( 0\right ) =0\)  Then (8) becomes

\begin{align*} 0 & =c_{1}\lim _{x\rightarrow 0}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +\left ( \frac {1}{x^{2}}+\frac {1}{4}-\frac {1}{288}x^{4}+\cdots \right ) \right ) \\ & =c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +\left ( \frac {1}{x^{2}}+\frac {1}{4}-\frac {1}{288}x^{4}+\cdots \right ) \right ) \end{align*}

We see the second solution is not defined at \(x=0\). Hence we let \(c_{2}=0\) and \(c_{1}\) remains arbitrary. So the solution (8) becomes

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \]

IC \(y\left ( 0\right ) =1\)  Then (8) becomes

\begin{align*} 1 & =c_{1}\lim _{x\rightarrow 0}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +\left ( \frac {1}{x^{2}}+\frac {1}{4}-\frac {1}{288}x^{4}+\cdots \right ) \right ) \\ 1 & =c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +\left ( \frac {1}{x^{2}}+\frac {1}{4}-\frac {1}{288}x^{4}+\cdots \right ) \right ) \end{align*}

We see the second solution is not defined at \(x=0\). Hence we let \(c_{2}=0\) and the above gives

\[ 1=0 \]

Hence no solution exist with this initial conditions.

IC \(y^{\prime }\left ( 0\right ) =0\)  Taking derivative of (8) gives

\[ y^{\prime }=c_{1}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) +c_{2}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \]

Then at \(x=0\) the above gives

\begin{align*} 0 & =c_{1}\lim _{x\rightarrow 0}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \\ & =c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \end{align*}

Since limit is not defined, we set \(c_{2}=0\). Therefore the solution (8) becomes

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \]

IC \(y^{\prime }\left ( 0\right ) =1\)  Taking derivative of (8) gives

\[ y^{\prime }=c_{1}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) +c_{2}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \]

Then at \(x=0\) the above gives

\begin{align*} 1 & =c_{1}\lim _{x\rightarrow 0}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \\ 1 & =c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \ln \left ( x\right ) +\frac {1}{x}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {-1}{x^{3}}-\frac {4}{288}x^{3}+\cdots \right ) \right ) \end{align*}

Since limit is not defined, we set \(c_{2}=0\) and the above becomes

\[ 1=0 \]

Hence no solution exist with this initial conditions.

IC \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0\)

Using \(y\left ( 0\right ) =0\) the solution (8) becomes

\[ 0=c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +x^{-2}\left ( 1+\frac {1}{4}x^{2}-\frac {1}{288}x^{6}+\cdots \right ) \right ) \]

Since limit is not defined, we set \(c_{2}=0\) and the solution now becomes

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \]

Taking derivatives gives

\[ y^{\prime }=c_{1}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \]

At \(x=0\), using the second IC gives

\begin{align*} 0 & =c_{1}\lim _{x\rightarrow 0}\left ( 2x-\frac {4}{12}x^{3}+\frac {6}{384}x^{5}+\cdots \right ) \\ 0 & =c_{1}0 \end{align*}

Hence \(c_{1}\) is arbitrary. It can be any value. Therefore the solution is

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \]

Notice that even though we had two initial conditions, the final solution still has one arbitrary constant in it.