What is the solution for x^3+8y(dy)/(dx)=0 ?

The solution for x^3+8y(dy)/(dx)=0 is y=(sqrt(-x^4+c_{1)})/4 ,y=-(sqrt(-x^4+c_{1)})/4

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x^3+8y(dy)/(dx)=0

x^{3} + 8 y \frac{d y}{d x} = 0

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

Solution steps

x^{3} + 8 y \frac{d y}{d x} = 0

Solve separable ODE:

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

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

\frac{d y}{d x} = y (8 x^{2} + 1)

\frac{1}{y ^{2}} - \frac{3 x}{y ^{3}} \frac{d y}{d x} = 0

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

y^{2} y^{^{'}} = 2 (t + t y^{3}) e^{t^{2}}, y (0) = 0

y^{^{'}} (x) = \frac{2 x - 1}{3 y ^{2} - 3}, y (1) = 0

What is the solution for x^3+8y(dy)/(dx)=0 ?
The solution for x^3+8y(dy)/(dx)=0 is y=(sqrt(-x^4+c_{1)})/4 ,y=-(sqrt(-x^4+c_{1)})/4