Question

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

${\mathit{x}}^{\mathbf{3}}\left(x^2-25\right)\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

x = 0, irregularx = 2, - 5,5,regular

Step-by-step solution

To find the singular points, regular points, and irregular points.

For second order differential equations in the form;

$P\left(x\right)y\text{'}\text{'}+Q\left(x\right)y\text{'}+R\left(x\right)=0$

Singular points are the values of x that make p(x) equal Zero.

$P\left(x\right)=x^3\left(x^2-25\right)(x-2{)}^2=x^3(x-2{)}^2(x-5)(x+5)$

The singular point is$x=0,x=2,x=5,x=-5$

$Q\left(x\right)=3x(x-2), R\left(x\right)=7(x+5)$

To determine whether a singular point is regular or irregular, we calculate the limits below,

If both limits are finite, then the singular points are regular otherwise it is irregular.

$\underset{x\to x_0}{\lim }\left(x-x_0\right)\frac{Q\left(x\right)}{P\left(x\right)}\text{and}\underset{x\to x_0}{\lim }{\left(x-x_0\right)}^2\frac{R\left(x\right)}{P\left(x\right)}$

When x = 0,

$$\begin{gathered} \underset{x\to 0}{\lim }x\frac{3x(x-2)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to 0}{\lim }\frac{3}{x{(x-2)}^2(x-5)(x+5)} \\ =\frac{3}{0} \\ =\lim it doesn\text{'}t exist \end{gathered}$$

The limits do not exist, thus x = 0 the is an irregular singular point.

When x = 2,

$$\begin{gathered} \underset{x\to 2}{\lim }(x-2)\frac{3x(x-2)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to 2}{\lim }\frac{3}{x^2(x-5)(x+5)} \\ =-\frac{1}{28} \\ \underset{x\to 2}{\lim }(x-2{)}^2\frac{7(x+5)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to 2}{\lim }\frac{7}{x^3(x-5)} \\ =-\frac{7}{24} \end{gathered}$$

Finite then x = 2 is a regular singular point.

Whenx = -5,

$$\begin{gathered} \underset{x\to -5}{\lim }(x+5)\frac{3x(x-2)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to -5}{\lim }\frac{3}{x^2(x-2)(x-5)} \\ =\frac{3}{1750} \end{gathered}$$

$$\begin{gathered} \underset{x\to -5}{\lim }(x+5{)}^2\frac{7(x+5)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to -5}{\lim }\frac{7{(x+5)}^2}{x^3{(x-2)}^2(x-5)} \\ =0 \end{gathered}$$

Then x = -5 is a regular singular point.

$$\begin{gathered} \underset{x\to 5}{\lim }(x-5)\frac{3x(x-2)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to 5}{\lim }\frac{3}{x^2(x-2)(x+5)} \\ =\frac{7}{250} \end{gathered}$$

$$\begin{gathered} \underset{x\to 5}{\lim }(x-5{)}^2\frac{7(x-5)}{x^3{(x-2)}^2(x-5)(x+5)}=\underset{x\to 5}{\lim }\frac{7(x-5)}{x^3{(x-2)}^2} \\ =0 \end{gathered}$$

Final result.

Thus,x = 0irregular, and x = 2,-5,5 are regular.