Answer to Question #289536 in Differential Equations for Ramya

Question #289536

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

1
Expert's answer
2022-01-24T16:50:11-0500

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


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