Question

In Problems 11 and 12 put the given differential equation into form (3) for each regular singular point of the equation. Identify the functions $\mathit{p}\left(x\right)$and $\mathit{q}\left(x\right)$.

role="math" localid="1664094425610" $\mathbf{11}\left(x^2-1\right)\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\left(x+1\right){\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\left(x^2-x\right)\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

For$x=1,{\left(x-1\right)}^{2}y\text{'}\text{'}+5\left(x-1\right)y^{\text{'}}+\frac{x{\left(x-1\right)}^2}{\left(x+1\right)},y=0,p\left(x\right)=5$ and$q\left(x\right)=\frac{x{\left(x-1\right)}^2}{\left(x+1\right)}$

For$x=-1,{\left(x+1\right)}^{2}y\text{'}\text{'}+\left(x+1\right)\frac{5\left(x+1\right)}{\left(x-1\right)}{y}^{\text{'}}+x\left(x+1\right),y=0,p\left(x\right)=\frac{5\left(x+1\right)}{\left(x-1\right)}$ and$q\left(x\right)=x\left(x+1\right)$

Step-by-step solution

To Find differential equation

We have

$\left(x^2-1\right)y\text{'}\text{'}+5(x+1)y^{\text{'}}+\left(x^2-x\right)y=0$

$\left(x-1\right)\left(x+1\right)y\text{'}\text{'}+5\left(x+1\right)y^{\text{'}}+x\left(x-1\right)y=0$

Our aim is to put the given differential equation in the form of (3), in the book, which is

${\left(x-x_0\right)}^{2}y\text{'}\text{'}+\left(x-x_0\right)p\left(x\right)y^{\text{'}}+q\left(x\right)y=0$

Where is a regular singular point. We need first to find the singular points of the given differential equation and classify it into regular points and irregular points. First, let's assume we have a differential equation with the following form

$$\begin{gathered} a_1\left(x\right)y\text{'}\text{'}+a_2\left(x\right)y^{\text{'}}+a_3\left(x\right)y=0 \\ y\text{'}\text{'}+\frac{a_2\left(x\right)}{a_1\left(x\right)}{y}^{\text{'}}+\frac{a_3\left(x\right)}{a_2\left(x\right)}y=0\left[\text{Dividing}\text{by}{\text{a}}_{\text{1}}\left(\text{x}\right)\right] \\ y\text{'}\text{'}+P\left(x\right)y^{\text{'}}+Q\left(x\right)y=0\dots \left(1\right) \end{gathered}$$

where

$P\left(x\right)=\frac{a_2\left(x\right)}{a_1\left(x\right)}\text{and}Q\left(x\right)=\frac{a_3\left(x\right)}{a_1\left(x\right)}\dots$

Equation (l) is the standard form, upon which we will build our analysis. We recall the following definition.

We break down the definition into the following; If you have an analytic function at a point, this implies that you can expand this function around that point using Talyor Series, which is calculated by the function's derivatives at that point. If the function has singular points, you will not be able to find the function's derivatives at that point, and it will blow up. Investigating (2), you will be able to deduce that and are not analytic at the points, which will make the denominator equals to zero, after common factors reduction. Now, we put the given differential equation in the standard form.

$$\begin{gathered} \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}y\text{'}\text{'}+\frac{5\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}{y}^{\text{'}}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}y=0 \\ y\text{'}\text{'}+\frac{5}{\left(x-1\right)}{y}^{\text{'}}+\frac{x}{\left(x+1\right)}y=0 \end{gathered}$$

yields,

$P\left(x\right)=\frac{5}{\left(x-1\right)}\text{and}Q\left(x\right)=\frac{x}{\left(x+1\right)}\dots \left(3\right)$

Now, we equate the denominator by 0 and find the values which satisfies this condition.

For P(x)

$x-1=0\Rightarrow x=1$

For Q(x)

$x+1=0\Rightarrow x=-1$

Therefore, P(x) and Q(x) are not analytic at$x=2$and $x=-3$, which implies that the differential equation has a singular points at$x=2$and $x=-3$. Now, we need to classify these singular points. We define two functions as follows:

$p\left(x\right)=\left(x-x_0\right)P\left(x\right)\text{and}q\left(x\right)={\left(x-x_0\right)}^{2}Q\left(x\right)\dots$

Where$x_{\circ }$is the singular point. Now, Recall the following definition.

We are left with classifying each point, at which the differential equation has a singular point. There is a way, by which we can classify the singular points, by taking the limit for (4) as goes x to $x_{\circ }$. If the limit exists, then the function is analytic at that point. And if the limit doesn't exist, this implies that the function is not analytic at that point.

At $x_{\circ }$, we find p(x) and q(x) as follows. From (3) and (4), we get

$$\begin{gathered} \underset{x\to 1}{\lim }p\left(x\right)=\underset{x\to 1}{\lim }(x-1)\times \frac{5}{(x-1)} \\ =5\left[p\right(x)isanalyticatx=1] \\ \underset{x\to 1}{\lim }q\left(x\right)=\underset{x\to 1}{\lim }(x-1{)}^2\times \frac{x}{x+1} \\ =0\left[q\right(x)isanalyticatx=1] \end{gathered}$$

Therefore, x=1 is a Regular Singular Point.

At x=-1

From (3) and (4), we get

$$\begin{gathered} \underset{x\to -1}{\lim }p\left(x\right)=\underset{x\to -1}{\lim }(x+1)·\frac{5}{(x-1)} \\ =0\left[\text{p}\right(\text{x})\text{is analytic at x}=-1] \\ \underset{x\to -1}{\lim }q\left(x\right)=\underset{x\to -1}{\lim }(x+1{)}^2·\frac{x}{x+1} \\ =\underset{x\to -1}{\lim }x(x+1) \\ =0\left[\text{q}\right(\text{x})\text{is analytic at x}=-1] \end{gathered}$$

Therefore, x=-1 and x=1 are Regular Singular Points. Now, we put the given differential equation in the form of (3), for the two regular singular points.

The Regular Singular Point

Therefore,$x=1$is a Regular Singular Point.

At$x=-1$

From (3) and (4), we get

$$\begin{gathered} \underset{x\to -1}{\lim }p\left(x\right)=\underset{x\to -1}{\lim }(x+1)·\frac{5}{(x-1)} \\ =0\left[\text{p}\right(\text{x})\text{is analytic at x}=-1] \\ \underset{x\to -1}{\lim }q\left(x\right)=\underset{x\to -1}{\lim }(x+1{)}^2·\frac{x}{x+1} \\ =\underset{x\to -1}{\lim }x(x+1) \\ =0\left[\text{q}\right(\text{x})\text{is analytic at x}=-1] \end{gathered}$$

Therefore,$x=-1$and$x=1$are Regular Singular Points. Now, we put the given differential equation in the form of (3), for the two regular singular points.

For$x=1$

We have

$$\begin{gathered} p\left(x\right)=(x-1)\frac{5}{(x-1)} \\ =5 \\ q\left(x\right)={(x-1)}^2\frac{x}{(x+1)} \\ =\frac{x{(x-1)}^2}{(x+1)} \end{gathered}$$

And the differential equation has the form

${\left(x-x_0\right)}^{2}y\text{'}\text{'}+\left(x-x_0\right)p\left(x\right)y^{\text{'}}+q\left(x\right)y=0$

$(x-1{)}^{2}y\text{'}\text{'}+5(x-1)y\text{'}+\frac{x{(x-1)}^2}{(x+1)}y=0$

Final Answer

For $x=-1$,

We have

$$\begin{gathered} p\left(x\right)=(x+1)\frac{5}{(x-1)} \\ =\frac{5(x+1)}{(x-1)} \\ q\left(x\right)={(x+1)}^2\frac{x}{(x+1)} \\ =x(x+1) \end{gathered}$$

And the differential equation has the form:

$$\begin{gathered} {\left(x-x_0\right)}^{2}y\text{'}\text{'}+\left(x-x_0\right)p\left(x\right)y^{\text{'}}+q\left(x\right)y=0 \\ (x+1{)}^{2}y\text{'}\text{'}+(x+1)\frac{5(x+1)}{(x-1)}{y}^{\text{'}}+x(x+1)y=0 \end{gathered}$$

The answer is:

For$x=1,{\left(x-1\right)}^{2}y\text{'}\text{'}+5\left(x-1\right)y^{\text{'}}+\frac{x{\left(x-1\right)}^2}{\left(x+1\right)},y=0,p\left(x\right)=5$ and$q\left(x\right)=\frac{x{\left(x-1\right)}^2}{\left(x+1\right)}$

For$x=-1,{\left(x+1\right)}^{2}y\text{'}\text{'}+\left(x+1\right)\frac{5\left(x+1\right)}{\left(x-1\right)}{y}^{\text{'}}+x\left(x+1\right),y=0,p\left(x\right)=\frac{5\left(x+1\right)}{\left(x-1\right)}$ and$q\left(x\right)=x\left(x+1\right)$