4.1 Example 1. homogeneous ode \(x^{2}y^{\prime \prime }+x^{2}y^{\prime }+y=0\)

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

With expansion around \(x=0\). This is a regular singular ODE. Let

\[ y\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n+r}\]

Where \(r\) is to be determined. It is the root of the indicial equation. Therefore

\begin{align*} y^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime }\left ( x\right ) & =\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^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\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+1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =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 middle term above as follows

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

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

\(n=0\) gives the indicial equation

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

EQ (3B) is used to solve for \(r\). Since \(a_{0}\neq 0\) then (3B) gives

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

The roots are

\begin{align*} r_{1} & =\frac {1}{2}+\frac {1}{2}i\sqrt {3}\\ r_{2} & =\frac {1}{2}-\frac {1}{2}i\sqrt {3}\end{align*}

The roots will always be complex conjugate of each others (since second order ode) and the real part will always be equal. Let the roots be

\[ r_{1,2}=\alpha \pm i\beta \]

When this happens, the solution is given similar to the case when the roots differ by non integer, except now the solution and the coefficients will be complex. Let the solution be

\[ y\left ( x\right ) =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \]

Now \(y_{1}\left ( x\right ) \) is solved for. The solution is

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

Starting with Eq (2A) which was derived above gives

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

The case of \(n=0\) is skipped since this was used to find the roots and \(a_{0}\neq 0.\) \(n\geq 1\) gives the recursion equation

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

For \(n=1\) Eq. (3) gives

\[ a_{1}=\frac {-r_{1}a_{n-1}}{\left ( 1+r_{1}\right ) \left ( r_{1}\right ) +1}\]

But \(r_{1}=\alpha \pm i\beta =\frac {1}{2}+\frac {1}{2}i\sqrt {3}\). and \(a_{0}=1\). Hence the above becomes

\[ a_{1}=\frac {-\left ( \frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) }{\left ( 1+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) \left ( \frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) +1}=\frac {-1}{2}\]

For \(n=2\) Eq. (3) gives

\[ a_{2}=\frac {-\left ( 1+r_{1}\right ) a_{1}}{\left ( 2+r_{1}\right ) \left ( 1+r_{1}\right ) +1}=\frac {-\left ( 1+r_{1}\right ) \left ( \frac {-1}{2}\right ) }{\left ( 2+r_{1}\right ) \left ( 1+r_{1}\right ) +1}=\frac {-\left ( 1+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) \left ( \frac {-1}{2}\right ) }{\left ( 2+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) \left ( 1+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) +1}=\frac {9}{56}-\frac {1}{56}i\sqrt {3}\]

For \(n=3\) Eq. (3) gives

\[ a_{3}=\frac {-\left ( 3+r_{1}-1\right ) a_{2}}{\left ( 3+r_{1}\right ) \left ( 2+r_{1}\right ) +1}=\frac {-\left ( 3+r_{1}-1\right ) \left ( \frac {9}{56}-\frac {1}{56}i\sqrt {3}\right ) }{\left ( 3+r_{1}\right ) \left ( 2+r_{1}\right ) +1}=\frac {-\left ( 2+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) \left ( \frac {9}{56}-\frac {1}{56}i\sqrt {3}\right ) }{\left ( 3+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) \left ( 2+\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right ) +1}=\frac {1}{112}i\sqrt {3}-\frac {13}{336}\]

And so on. Hence the first solution is

\begin{equation} y_{1}\left ( x\right ) =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}=x^{\frac {1}{2}+\frac {1}{2}i\sqrt {3}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \tag {4}\end{equation}

but

\[ x^{\frac {1}{2}+\frac {1}{2}i\sqrt {3}}=x^{\frac {1}{2}}x^{\frac {1}{2}i\sqrt {3}}=x^{\frac {1}{2}}e^{\ln \left ( x^{\frac {1}{2}i\sqrt {3}}\right ) }=x^{\frac {1}{2}}e^{i\ln \left ( x^{\frac {\sqrt {3}}{2}}\right ) }=x^{\frac {1}{2}}\left ( \cos \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) +i\sin \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) \right ) \]

Substituting the above in (4) and using values found for \(a_{n}\) gives

\begin{equation} y_{1}\left ( x\right ) =x^{\frac {1}{2}}\left ( \cos \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) +i\sin \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) \right ) \left ( 1-\frac {1}{2}x+\left ( \frac {9}{56}-\frac {1}{56}i\sqrt {3}\right ) x^{2}+\left ( -\frac {13}{336}+\frac {1}{112}i\sqrt {3}\right ) x^{3}+\cdots \right ) \tag {5}\end{equation}

Since the roots are complex conjugate of each others, then the second solution is

\begin{equation} y_{2}\left ( x\right ) =x^{\frac {1}{2}}\left ( \cos \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) -i\sin \left ( \ln x^{\frac {\sqrt {3}}{2}}\right ) \right ) \left ( 1-\frac {1}{2}x+\left ( \frac {9}{56}+\frac {1}{56}i\sqrt {3}\right ) x^{2}+\left ( -\frac {13}{336}-\frac {1}{112}i\sqrt {3}\right ) x^{3}+\cdots \right ) \tag {6}\end{equation}

The final solution is therefore

\[ y\left ( x\right ) =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \]