What is the solution for xyy^'+y^2=2x ?

The solution for xyy^'+y^2=2x is y=(sqrt(4x^3+c_{1)})/(sqrt(3)x),y=-(sqrt(4x^3+c_{1)})/(sqrt(3)x)

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xyy^'+y^2=2x

x y y^{^{'}} + y^{2} = 2 x

y = \frac{4 x ^{3} + c _{1}}{3 x}, y = - \frac{4 x ^{3} + c _{1}}{3 x}

Solution steps

x y y^{'} (x) + y^{2} = 2 x

Solve Bernoulli ODE:

y = \frac{4 x ^{3} + c _{1}}{3 x}, y = - \frac{4 x ^{3} + c _{1}}{3 x}

y = \frac{4 x ^{3} + c _{1}}{3 x}, y = - \frac{4 x ^{3} + c _{1}}{3 x}

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What is the solution for xyy^'+y^2=2x ?
The solution for xyy^'+y^2=2x is y=(sqrt(4x^3+c_{1)})/(sqrt(3)x),y=-(sqrt(4x^3+c_{1)})/(sqrt(3)x)