I have first-order differential equation $$y=xy'+ \frac{1}{2}(y')^{2}$$ Maybe, with this someone will find way to solve it $$\frac{1}{2}y'(2x+y')=y$$ I thought I can use $x^2+y=t$ for subtitution and when I derivate, I have $t'=2x+y'\\(t'-2x)t'=2t-2x^2$ which is acctualy the same as previous. I don't have idea how to start..
Help solving first-order differential equation
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ordinary-differential-equations
2 Answers
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Hint: Rewrite like this: $$2y = 2xy'+y'^2$$ so adding $x^2$ on both sides we get $$2y+x^2 = x^2+2xy'+y'^2$$ or $$2y+x^2 = (x+y')^2$$
Let $z = 2y+x^2$ then $z' = 2y'+2x$, so $$4z =(z')^2\implies z'=\pm 2\sqrt{z} \implies z = (c\pm x)^2$$
So $$ y = {1\over 2}(z-x^2) = {1\over 2}c^2\pm cx$$
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Differentiate both sides with respect to $x$: $$y'=xy''+y'+y'y''$$ and then we get $$y''(x+y')=0$$