(d^2 y/dx) +x (dy/dx) −y=0


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Question Number 120058 by bramlexs22 last updated on 29/Oct/20

$\frac{d^{2} y}{d x} + x \frac{d y}{d x} - y = 0$
	Answered by Olaf last updated on 29/Oct/20

	$y^{″} + {x y}^{'} - y = 0 (1)$ $y = x u$ $y^{'} = {x u}^{'} + u$ $y^{″} = {x u}^{″} + 2 u^{'}$ $(1) : {x u}^{″} + 2 u^{'} + x ({x u}^{'} + u) - x u = 0$ ${x u}^{″} + 2 u^{'} + x^{2} u^{'} = 0$ $\frac{u^{″}}{u^{'}} = - \frac{2 + x^{2}}{x} = - \frac{2}{x} - x$ $\ln ∣ u^{'} ∣ = - 2 \ln ∣ x ∣ - \frac{x^{2}}{2} + C_{1}$ $u^{'} = C_{2} e^{\ln \frac{1}{x^{2}}} e^{- \frac{x^{2}}{2}} = \frac{C_{2}}{x^{2}} e^{- \frac{x^{2}}{2}}$ $u = - C_{2} [\frac{e^{- \frac{x^{2}}{2}}}{x} + \sqrt{\frac{π}{2}} \erf (\frac{x}{\sqrt{2}})] + C_{3}$ $(By parts)$ $y = u x = - C_{2} [e^{- \frac{x^{2}}{2}} + \sqrt{\frac{π}{2}} x \erf (\frac{x}{\sqrt{2}})] + C_{3} x$
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