yy′′−(y′)^2 =0


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Question Number 126460 by BHOOPENDRA last updated on 20/Dec/20

${y y}^{″} - {(y^{'})}^{2} = 0$
Commented by BHOOPENDRA last updated on 20/Dec/20

$t h a n k s a l o t s i r$
	Answered by mr W last updated on 20/Dec/20

	$p = y^{'}$ $y^{″} = \frac{d p}{d x} = p \frac{d p}{d y}$ $y p \frac{d p}{d y} - p^{2} = 0$ $\Rightarrow p = 0 \Rightarrow y = C$ $\Rightarrow y \frac{d p}{d y} = p \Rightarrow \frac{d p}{p} = \frac{d y}{y} \Rightarrow \ln p = \ln C y$ $\Rightarrow p = C y \Rightarrow \frac{d y}{d x} = C y \Rightarrow \frac{d y}{y} = C d x$ $\Rightarrow \ln y = C x + C_{2}$ $\Rightarrow y = C_{1} e^{C x}$
	Commented by Coronavirus last updated on 20/Dec/20

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