Question

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular.

$\mathbf{(}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore $x=-2i$,$x=2i$, $x=0$and are regular singular points.

Step-by-step solution

To Find the singular points of the given differential equation

We have

$\left(x^3+4x\right)y\text{'}\text{'}-2xy\text{'}+6y=0$

We aim to find the singular points of the given differential equation and classify them. First, let's assume we have a differential equation with the following form;

$$\begin{gathered} a_1\left(x\right)y\text{'}\text{'}+a_2\left(x\right)y\text{'}+a_3\left(x\right)y=0 \\ y\text{'}\text{'}+\frac{a_2\left(x\right)}{a_1\left(x\right)}y\text{'}+\frac{a_3\left(x\right)}{a_2\left(x\right)}y=0 \\ y\text{'}\text{'}+P\left(x\right)y\text{'}+Q\left(x\right)y=0 \dots \left(1\right) \end{gathered}$$

Where;

$P\left(x\right)=\frac{a_2\left(x\right)}{a_1\left(x\right)}andQ\left(x\right)=\frac{a_3\left(x\right)}{a_1\left(x\right)}\dots \left(2\right)$

Find the ordinary points of the differential equation.

A point $x=x_0$is said to be an ordinary point of the differential equation, if we have an analytic function at a point, this implies that we can expand this function around that point using Taylor 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.

We will be able to deduce that P(x) and Q(x)are not analytic at the points, which will make the denominator equal to zero, after common factors reduction. Now, we put the given differential equation into the standard form.

$$\begin{gathered} \frac{x\left(x^2+4\right)}{x\left(x^2+4\right)}y"-\frac{2x}{x\left(x^2+4\right)}y\text{'}+\frac{6}{x\left(x^2+4\right)}y=0 \\ y"-\frac{2}{\left(x^2+4\right)}y\text{'}\frac{6}{x\left(x^2+4\right)}y=0 \end{gathered}$$

Therefore, we have:

$P\left(x\right)=-\frac{2}{\left(x^2+4\right)}andQ\left(x\right)=\frac{6}{x\left(x^2+4\right)}$

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

For P(x);

$x^2+4=0\Rightarrow x=\pm 2i$

For Q(x);

$x\left(x^2+4\right)=0\Rightarrow x=\left\{-2i,2i,0\right\}$

Therefore, P(x) and Q(x) are not analytic at $x=2i,x=-2i$ and x = 0 which implies that the differential equation has a singular point at $x=2i,x=-2i$and x = 0. Now, we need to classify these singular points. We define two functions as follows:

$p\left(x\right)=\left(x-x_0\right)P\left(x\right)andq\left(x\right)={\left(x-x_0\right)}^{2}Q\left(x\right) \dots \left(4\right)$

Where x = 0 is the singular point.

A singular point $x_0,$is said to be a regular singular point of the differential equation, if P(x) and Q(x) are analytic at $x_0,$otherwise it is an irregular singular point.

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.

Final proof

At x = 0

$$\begin{gathered} \underset{x\to 0}{\lim }p\left(x\right)=\underset{x\to 0}{\lim }(x-0)·\frac{-2}{\left(x^2+4\right)} \\ =0 \left[\text{p}\right(\text{x})is analyticat \text{x}=0] \\ \underset{x\to 0}{\lim }q\left(x\right)=\underset{x\to 0}{\lim }{(x-0)}^2·\frac{6}{x\left(x2+4\right)}=\underset{x\to 0}{\lim }\frac{6x}{\left(x^2+4\right)} \\ =0 \left[\text{q}\right(\text{x})is analyticat \text{x}=0] \end{gathered}$$

Therefore, x = 0 is a regular singular point.

At $x=2i$

$$\begin{gathered} \underset{x\to 2i}{\lim }p\left(x\right)=\underset{x\to 2i}{\lim }(x-2i)·\frac{-2}{\left(x^2-{\left(2i\right)}^2\right)} \\ =\underset{x\to 2i}{\lim }(x-2i)·\frac{-2}{(x-2i)(x+2i)} \\ =\underset{x\to 2i}{\lim }\frac{-2}{(x+2i)} \\ =\frac{-1}{2i} \left[\text{p}\right(\text{x})is analyticat \text{x}=2i] \\ \underset{x\to 2i}{\lim }q\left(x\right)=\underset{x\to 2i}{\lim }{(x-2i)}^2·\frac{6}{x\left(x^{}+2i\right)(x-2i)} \\ =\underset{x\to 2i}{\lim }\frac{6(x-2i)}{x(x+2i)} \\ =0 \end{gathered}$$

Therefore, x = 2i is a Regular Singular point.

At x = -2i ,from (3) and (4), we get:

$$\begin{gathered} \underset{x\to 2i}{\lim }p\left(x\right)=\underset{x\to 2i}{\lim }(x+2i)·\frac{-2}{\left(x^2-{\left(2i\right)}^2\right)} \\ =\underset{x\to 2i}{\lim }(x+2i)·\frac{-2}{(x-2i)(x+2i)} \\ =\frac{1}{2i} \left[\text{p}\right(\text{x})is analyticat \text{x}=-2i] \\ \underset{x\to -2i}{\lim }q\left(x\right)=\underset{x\to -2i}{\lim }{(x+2i)}^2·\frac{6}{x\left(x^{}+2i\right)(x-2i)} \\ =\underset{x\to -2i}{\lim }\frac{6(x+2i)}{x(x-2i)} \\ =0 \left[\text{q}\right(\text{x}) is analyticat \text{x}=-2\text{i}] \end{gathered}$$

Therefore, x = -2i is a regular singular point.

Thus,x = 2i ,x = -2i and x = 0 are regular singular points.