Answer to Question #289536 in Differential Equations for Ramya

Question

For the following differential equation locate and classify its singular points on the x-axis x^2 y" + (2-x)y' = 0.

Answer 1

$\text{$x^2$y$\prime\prime+(2-x)$y$\prime=0$}\cdots\cdots\cdots(\text{eqn}1)\\ \text{Comparing eqn1 to the general equation below:}\\P(x)y\prime\prime+Q(x)y\prime+R(x)y=0 \text{, yields}\\ P=x^2, Q(x)=(2-x), \text{and }R(x)=0\\ \text{Now, by definition we have that a point }x_o \text{ is a singular point if} \ P(x_0)=0.\\ \text{Thus equating our }P(x)=0,\text{yields}\\ x^2=0 \Rightarrow x=0.\\ \text{The singular points of the given differential equation on the x-axis is \textbf{x\ =\ 0}}.\\ \text{To classify the singular point }x_o=0\text{, we need to compute,}\\ \lim_{x\rightarrow x_0}(x-x_o)\frac{Q(x)}{P(x)}, \text{and }\lim_{x\rightarrow x_0}(x-x_o)^2\frac{R(x)}{P(x)}\\ \text{Now, since }x_o=0,\text{ we have:}\\ \lim_{x\rightarrow x_0}(x-x_o)\frac{Q(x)}{P(x)}=\lim_{x\rightarrow 0}(x-0)(\frac{2-x}{x^2})=\lim_{x\rightarrow0}\frac{2-x}{x}=\infty, \text{and}\\ \lim_{x\rightarrow x_0}(x-x_o)^2\frac{R(x)}{P(x)}=0<\infty.\\ \text{ Since, }\lim_{x\rightarrow x_0}(x-x_o)^2\frac{R(x)}{P(x)}<\infty \text{ but}\\\text{ } \lim_{x\rightarrow x_0}(x-x_o)\frac{Q(x)}{P(x)}=\infty,\\ \text{We conclude that the singular point }x_o=0 \text{ is an \textbf{irregular singular point}.}$