1.1.3.1 Example 1. \(xy^{\prime }+y=0,y\left ( 0\right ) =1\)

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

With expansion around \(x=0\). Since \(x\) is regular singular point, then 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}\end{align*}

Then (1) becomes

\begin{align} x\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 {2}\\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \tag {3A}\end{align}

All \(x\) in sums start from same index, so there is no adjustments needed. For \(n=0\), EQ (3) gives

\begin{align} ra_{0}x^{r}+a_{0}x^{r} & =0\tag {*}\\ \left ( r+1\right ) a_{0}x^{r} & =0\nonumber \end{align}

Eq (*) above is important. It is called balance equation and will be used to find \(c_{0},r\) for particular solution for ode when the RHS is not zero. In this case it will not be needed but will be used in next problems. Eq (3A) is also very important. It will be used also to find particular solution when right side is not zero. From now will call this the summation equation. It is the final equation when all sums have been normalized such that power of \(x\) inside each sum all have same power.  Both the balance equation and the summation equation are the core of the algorithm for solving first order series equation with regular singular point and will be used over and over in all the problems.

Now we start by finding \(y_{h}\). From balance equation and since \(a_{0}\neq 0\), hence \(r+1=0\) or

\[ r=-1 \]

Therefore summation equation becomes

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

For \(n>0\) (because \(n=0\) was used to find \(r\)) EQ (3B) gives the recursive relation

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

We see that for all \(n>0\) then \(a_{n}=0\). Hence solution is for \(n=0\) only, which is

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ & =a_{0}x^{-1}\\ & =\frac {a_{0}}{x}\end{align*}

Initial condition now applied. We see that at \(x=0\) we get undefined value. Hence not possible to solve for \(a_{0}\). Therefore no solution exist with this initial condition.