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Find two-parameter families of solutions for the following differential equation $$ yy'' + (y')^2 = 1. $$

Can this be explained on the appropriote steps required to find the solution of $y = \pm \sqrt{x^2+Fx+G}$, where $F$ and $G$ are constants.

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The key observation is the fact that $$\dfrac{d}{dx}\left( y \dfrac{dy}{dx}\right) = y \dfrac{d^2 y}{dx^2} + \left( \dfrac{dy}{dx}\right)^2$$ Hence, we have that $$\dfrac{d}{dx}\left( y \dfrac{dy}{dx}\right) = 1$$ i.e. $$y \dfrac{dy}{dx} = x + c_1 \implies y^2 = x^2 + 2c_1x + c_2 \implies y = \pm \sqrt{x^2 + Fx + G}$$

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For this OP I would probably add $$ \dfrac{d}{dx}\left( y^2 \right) = 2 y \dfrac{dy}{dx} = 2 x + 2 c_1 $$ – Will Jagy Nov 18 '12 at 20:53

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