What is the solution for 2yy^'=6x^2 ?

The solution for 2yy^'=6x^2 is y=sqrt(2x^3+c_{1)},y=-sqrt(2x^3+c_{1)}

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2yy^'=6x^2

2 y y^{^{'}} = 6 x^{2}

y = 2 x^{3} + c_{1}, y = - 2 x^{3} + c_{1}

Solution steps

2 y y^{'} (x) = 6 x^{2}

Solve separable ODE:

y = 2 x^{3} + c_{1}, y = - 2 x^{3} + c_{1}

y = 2 x^{3} + c_{1}, y = - 2 x^{3} + c_{1}

e^{x} y \frac{d y}{d x} = e^{- 2 x - y}

7 x y \frac{d y}{d x} = 2, y > 0, x > 0

x^{^{'}} + 2 x = 12

1 - y^{2} + (1 + x^{2}) \cdot \frac{d y}{d x} = 0

\frac{d y}{d x} y = \frac{- 1}{x}

What is the solution for 2yy^'=6x^2 ?
The solution for 2yy^'=6x^2 is y=sqrt(2x^3+c_{1)},y=-sqrt(2x^3+c_{1)}