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Question Number 86294 by jagoll last updated on 28/Mar/20

$$\mathrm{y}=2\mathrm{x}+{\left({\mathrm{y}}^{\prime }\right)}^2-{4\mathrm{y}}^{\prime }$$

Answered by mr W last updated on 28/Mar/20

$$y^{\prime }=2\pm \sqrt{4-2x+y}$$$$letu=4-2x+y$$$$\frac{du}{dx}=-2+\frac{dy}{dx}$$$$2+\frac{du}{dx}=2\pm \sqrt{u}$$$$\frac{du}{dx}=\pm \sqrt{u}$$$$\int \frac{du}{\sqrt{u}}=\pm \int dx$$$$2\sqrt{u}=\pm x+C$$$$\Rightarrow 2\sqrt{4-2x+y}\pm x=C$$$$\Rightarrow y=\frac{{(C\pm x)}^2}{4}+2x-4$$

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