prove that (y−c)^2 =cx its solution of the differention equation 4xy^(′′) +2xy^′ −y=0 (m.o)


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Question Number 115829 by mohammad17 last updated on 28/Sep/20

$p r o v e t h a t {(y - c)}^{2} = c x i t s s o l u t i o n o f t h e d i f f e r e n t i o n e q u a t i o n 4 {x y}^{^{″}} + 2 {x y}^{^{'}} - y = 0$ $(m . o)$
	Answered by $@y@m last updated on 29/Sep/20

	$y - c = \sqrt{c x}$ $y = \sqrt{c} . \sqrt{x} + c . . . (1)$ $y^{'} = \sqrt{c} . \frac{1}{2 \sqrt{x}} . . . (2)$ $y^{″} = \frac{\sqrt{c}}{2} . \frac{- 1}{2 x \sqrt{x}}$ $y^{″} = - \frac{\sqrt{c}}{4} . \frac{1}{x \sqrt{x}} . . . (3)$ $N o w,$ $4 {x y}^{^{″}} + 2 {x y}^{^{'}} - y = 4 x \times (- \frac{\sqrt{c}}{4} . \frac{1}{x \sqrt{x}}) + 2 x \times \frac{\sqrt{c}}{2 \sqrt{x}} - \sqrt{c x} - c$ $= \frac{- \sqrt{c}}{x \sqrt{x}} + \sqrt{c x} - . \sqrt{c x} - c$ $= \frac{- \sqrt{c}}{x \sqrt{x}} - c$ $\neq 0$ $P l . c h e c k q u e s t i o n$ $O R, m y s o l u t i o n .$ $:)$
	Commented by mohammad17 last updated on 29/Sep/20

	$t h a n k y o u s i r$
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