It was explained in the last
chapter that we have to analyse first whether the point is ordinary or
singular. In the case the point is *ordinary*, we can find solution around that
point by power series. The solution around singular points has been left to explain.

For **example** DE

has two singular points 0 and 1. If we try to find solution of DE at singular
points *by successive differentiation* in the form of power series

we would run into trouble. Since $y''(0)$ is not defined at singular point, also $c_2 = y''(0)/2!$ (from Taylor series) does not exist. Moreover, if we try to use method of undetermined coefficients to determine $a_0,\ a_1,\ \dots,$ they will be all zero. That is because after division by $x^4$ the coefficient of $y'$ becomes not analytic at $x=0$ and can not be expressed by Maclaurin series (we can not collect members of Maclaurin series at zero for a function $1/x$ or $1/x^3$.

**If power series for $\boldsymbol y$ is modified by method of Frobenius, then—at
regular singularity points—the solution can be expressed.** Now we owe to divide singular points into

**regular singular points**(solution can be found by Frobenius method) and**irregular singular points**(the problem of finding the series is too difficult to discuss here).

Let us simplify the discussion only to the second order differential equations in the standard form of

$$
\begin{equation}
y'' + P(x)y' + Q(x)y = 0. \label{ref:sing1}
\end{equation}
$$

Let us **multiply** functions

- $P(x)$ by $(x-x_0)$,
- $Q(x)$ by $(x-x_0)^2$.

After that,

- if
*each*member $P(x)\cdot(x-x_0)$ and $Q(x)\cdot(x-x_0)^2$ is analytic at $x=x_0$ then the singular point is regular. - Otherwise point is irregular singularity of DE.

Determine the singularities of DE $(x-1)^3x^2y''-2(x-1)xy'-3y=0$.

First let us bring DE into standard form $(\ref{ref:sing1})$.

$$ y''-\frac{2}{x(x-1)^2}y' - \frac{3}{(x-1)^3x^2}y = 0 $$Two singularities are observed. At $x=\{0, 1\}$.

- According to the rule, for the singularity at $x=0$ we have to multiply $P(x) = \frac{-2}{x(x-1)^2}$ by $x$ and $Q(x)=\frac{-3}{(x-1)^3x^2}$ by $x^2$. Then both members are analytic so the DE is a regular singularity at $x=0$.
- For the case at $x=1$, if we multiply $P(x)$ by $(x-1)$ and $Q(x)$ by $(x-1)^2$, these members do not become analytic. So at $x=1$ DE is irregular singularity.

Determine the singularities of DE $(x-1)^2x^4y'' + 2(x-1)xy'-y = 0$.

First let us bring DE into standard form $(\ref{ref:sing1})$.

$$ y'' + \frac{2}{(x-1)x^3}y' - \frac{1}{(x-1)^2x^4}y = 0 $$Two singularities are observed. At $x=\{0, 1\}$.

- According to the rule, for the singularity at $x=0$ we have to multiply $P(x) = \frac{2}{(x-1)x^3}$ by $x$ and $Q(x)=\frac{-1}{(x-1)^2x^4}$ by $x^2$. Then both members are still not analytic so the DE is a irregular singularity at $x=0$.
- For the case at $x=1$, if we multiply $P(x)$ by $(x-1)$ and $Q(x)$ by $(x-1)^2$, these members become analytic. So at $x=1$ DE is regular singularity.

**If $\boldsymbol{x=x_0}$ is a regular singularity then the series solution at such point will be found in the form of Frobenius series**

For $r=0$ or a positive integer the series $(\ref{ref:sing2})$ becomes Taylor series. A Frobenius therefore, includes the Taylor series as a special case.

When solving by means of Frobenius series, one has to deal with
**indicial equation** and its roots $r_1,\ r_2$. It will be demonstrated below
on example, that for the second order DE we have two roots to expect. Each root
is an origin of one of the solutions of DE then.

It might happen then that

**Roots are distinct, their difference not equal to an integer**(e.g. 1, 2, 3, ...)

Each root $r_1,\ r_2$ is an origin of a series forming a solution.**Roots are distinct, their difference equal to an integer**

In some cases each root is an origin of a series solution, in other cases there will be only one series $y_1(x)$ and the other solution $y_2(x)$ has to be determined as a logarithmic solution from $y_1(x)$ and such logarithmic solution has limited applications.**Roots are the same**

There will be only one series $y_1(x)$ and the other solution $y_2(x)$ has to be determined as a logarithmic solution from $y_1$ (and has limited applications).

Find solution of DE $x^2y''+x(x+\frac{1}{2})y'+xy=0$ at regular singularity.

First, if we bring DE into standard form $(\ref{ref:sing1})$ and apply rules to test singularities, it will be shown that $x=0$ is a regular singularity. We will solve given DE in $x=0$ by Frobenius series.

According to the Frobenius method, the solution will be found in the form

$$y = {(x-x_0)^r} \ [c_0 + c_1(x-x_0) + c_2 (x-x_0)^2 +c_3(x-x_0)^3 + \dots $$where two roots $r_1,\ r_2$ has to be evaluated. The coefficient $c_0$ is an arbitrary constant and the other constants will be expressed in terms of $c_0$. Thus we might expect two independent solutions.

Let us prepare $y,\ y',\ y''$ at $x_0=0$:

$$ \begin{align} & y = x^r(c_0+c_1x +c_2x^2 + c_3x^3 + \dots) \nonumber \\ & y' = c_0rx^{r-1} + c_1(r+1)x^r + c_2(r+2)x^{r+1}+\dots \nonumber \\ & y'' = c_0r(r-1)x^{r-2} + c_1(r+1)rx^{r-1} + c_2(r+2)(r+1)x^r+\dots \nonumber \end{align} $$Let us substitute $y,\ y',\ y''$ into given DE

$$ \begin{eqnarray} \color{blue}{(c_0r(r-1)x^r+c_1(r+1)rx^{r+1} + c_2(r+2)(r+1)x^{r+2}}&&\color{blue}+&& \color{blue}{\dots)} &&+&& \label{ref:sing3}\\ + (c_0rx^{r+1} + c_1(r+1)x^{r+2} + c_2(r+2)x^{r+3} + \dots &&+&& && && \label{ref:sing4}\\ + \frac 1 2 c_0 rx^r + \frac 1 2 c_1(r+1)x^{r+1} + \frac 1 2 c_2(r+2)x^{r+2} &&+&& \dots) &&+&& \label{ref:sing5}\\ + \color{green}{(c_0x^{r+1}+c_1x^{r+2} +c_2x^{r+3}} &&\color{green}+&& \color{green}{\dots)} &&=&& 0 \label{ref:sing6} \end{eqnarray} $$So above is given DE written be means of infinite series. On $(\ref{ref:sing3})$ is $x^2y''$, on $(\ref{ref:sing4})$ is $x^2y'$, on $(\ref{ref:sing5})$ is $\frac 1 2 xy'$, on $(\ref{ref:sing6})$ is $xy$.

Now we have to use the analogy method to that method of undetermined coefficients (note that for any $x^n$ there is always zero on the right side). Let us collect like powers of $x$ from the above equation:

$$ \begin{align} \color{red}{x^r}&(c_0r(r-1)+\frac 1 2 c_0r) = 0 \label{ref:sing7} \\ \color{red}{x^{r+1}}&(c_1(r+1)r + c_0r + \frac 1 2 c_1(r+1) + c_0) = 0 \label{ref:sing8} \\ \color{red}{x^{r+2}}&(c_2(r+2)(r+1) + c_1(r+1) + \frac 1 2 c_2(r+2) + c_1) = 0 \nonumber \\ \color{red}{x^{r+3}}&(c_3(r+3)(r+2) + c_2(r+2) + \frac 1 2 c_3(r+3) + c_2) = 0 \nonumber \end{align} $$Note that the first equation is used for a special purpose, i.e. to determine
roots $r$: the equation $(\ref{ref:sing7})$ can be rewritten to
${x^r}c_0(r(r-1)+\frac 1 2 r) = 0$ and is called **indicial
equation**. We have no
reason to expect that the first coefficients $c_0$ of series have to be zero.
Then to find $r_1,\ r_2$:

**The first solution** $y_1$ is based on $r=0$. From $(\ref{ref:sing8})$ we
can derive a recursion formula for coefficient $c_n$ when $r=0$:

And from the recursive rule forming coefficients it is possible to form solution $y_1$:

$$ {\underline{y_1(x) = c_0(1- 2x + \frac 4 3 x^2 - \frac 8 {15} x^3 + \dots)}} $$(If we use recursive formula of $c_n$, then e.g. for $c_0 = 1 \implies c_1 = -2$, for $c_1 = -2 \implies c_2 = \frac 4 3 $ and so on.)

**The second solution** $y_2$ is based on $r=1/2$. From $(\ref{ref:sing8})$ we
can derive a recursion formula for coefficient $c_n$ when $r=1/2$:

And if we use the last formula to determine coefficients $c_n$ of the second solution we get

$$ \underline{y_2(x) = c_0(1-x+\frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}- \dots)}. $$The general solution expressed as a power series is then a combination of $y_1$ and $y_2$.

$$ \underline{\underline{y(x) = C_0(1- 2x + \frac 4 3 x^2 - \frac 8 {15} x^3 + \dots) + C_1(1-x+\frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}- \dots)}} $$Find solution of DE $x^2y'' - 3xy' + 4(x+1)y = 0$ at regular singularity.

When the DE is converted into basic form, it can be observed that there is a regular singularity at $x=0$:

$$ y''-\frac{3y'}{x} + 4(\frac{x+1}{x^2})y = 0. $$We expect the solution $y$ in Frobenius form $(\ref{ref:sing2})$. Like in previous example, let us prepare $y',\ y''$

$$ \begin{align} y &= x^r(c_1+c_2x + c_3x^2 + c_4x^3 + \dots) \nonumber \\ y' &= rx^{r-1}c_1 + (r+1)x^rc_2 + (r+2)x^{r+1}c_3+ \dots \nonumber \\ y'' &= r(r-1)x^{r-2}c_1 + (r+1)rx^{r-1}c_2 + (r+2)(r+1)x^rc_3 + \dots \nonumber \\ \end{align} $$and substitute them into DE in basic form:

$$ \begin{align} &r(r-1)x^{r-2}c_1 + (r+1)rx^{r-1}c_2 + (r+2)(r+1)x^rc_3 + \dots - \nonumber \\ -&3\frac 1 x \left(rx^{r-1}c_1 + (r+1)x^rc_2 + (r+2)x^{r+1}c_3\right) \dots + \nonumber \\ +& 4\frac{1}{x}\left(x^r(c_1 +c_2x+c_3x^2 + c_4x^3)\right) + \nonumber \\ +& 4\frac 1 {x^2}\left(x^r(c_1+c_2x+c_3x^2+c_4x^3+\dots)\right) = 0 \nonumber \end{align} $$Now let us use the method of undetermined coefficients, first let us consider only the first possible term $x^{r-2}$ in order to find indicial equation and its roots $r_1,\ r_2:$

$$ \color{red}{x^{r-2}}\left[r(r-1)c_1 - 3rc_1 + 4c_1\right] = x^{r-2}\left[c_1(r^2 -4r + 4)\right] \implies r = \{2,\ 2\} $$The roots are the same. We will evaluate only the first solution (for $\boldsymbol{r_1=2}$), the other solution, which is logarithmic, we are going to leave unexpressed.

$$ \begin{align} \color{red}{x^{r-1}}\ &\left[(r+1)rc_2 - 3(r+1)c_2 + 4c_1 + 4c_2\right] = \nonumber \\ = x\ &\left[3\cdot 2 c_2 -3\cdot 3 c_2 + 4c_1 + 4c_2\right] \implies c_2 = -4c_1 \nonumber \\[3mm] \color{red}{x^r} \ &\left[(r+2)(r+1)c_3 - 3(r+2)c_3 + 4c_2+4c_3\right] = \nonumber \\ = x^2 \ &\left[4\cdot 3c_3 - 3c_3\cdot 4 + 4c_2+4c_3\right] \implies c_3 = -c_2 \nonumber \\[3mm] \color{red}{x^{r+1}} \ &\left[(r+3)(r+2)c_4 - 3(r+3)c_4 + 4c_3+4c_4\right] = \nonumber \\ = x^3 \ &\left[5\cdot 4c_4 - 3c_4\cdot 5 + 4c_3+4c_4\right] \implies c_4 = -\frac{4}{9}c_3 \nonumber \\ \end{align} $$And so on. Since we hav evaluated the coefficients, we are able to write the series:

$$ \underline{\underline{y= c_1x^2(1 - 4x + 4x^2 - \frac{16}{9}x^3 + \dots)}}. $$