What is the solution for y^'+2x^2y=(2x^2)/(sqrt(y)) ?

The solution for y^'+2x^2y=(2x^2)/(sqrt(y)) is y=\sqrt[3]{e^{-2x^3-3c_{1}}-2e^{-x^3-(3c_{1})/2}+1}

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y^'+2x^2y=(2x^2)/(sqrt(y))

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} + tan (3 x) y = 0, y (0) = 5

2 x y y^{^{'}} - 2 x^{2} + 1 = 0!

\frac{d y}{d x} = x y, y \neq = 0

y^{^{'}} = \frac{20 - 0.06 y}{0.02}

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

What is the solution for y^'+2x^2y=(2x^2)/(sqrt(y)) ?
The solution for y^'+2x^2y=(2x^2)/(sqrt(y)) is y=\sqrt[3]{e^{-2x^3-3c_{1}}-2e^{-x^3-(3c_{1})/2}+1}