What is the solution for x^2+3yy^'=0 ?

The solution for x^2+3yy^'=0 is y=(sqrt(-2x^3+c_{1)})/3 ,y=-(sqrt(-2x^3+c_{1)})/3

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x^2+3yy^'=0

x^{2} + 3 y y^{^{'}} = 0

y = \frac{- 2 x ^{3} + c _{1}}{3}, y = - \frac{- 2 x ^{3} + c _{1}}{3}

Solution steps

x^{2} + 3 y y^{'} (x) = 0

Solve separable ODE:

y = \frac{- 2 x ^{3} + c _{1}}{3}, y = - \frac{- 2 x ^{3} + c _{1}}{3}

y = \frac{- 2 x ^{3} + c _{1}}{3}, y = - \frac{- 2 x ^{3} + c _{1}}{3}

What is the solution for x^2+3yy^'=0 ?
The solution for x^2+3yy^'=0 is y=(sqrt(-2x^3+c_{1)})/3 ,y=-(sqrt(-2x^3+c_{1)})/3