What is the solution for 3y(dy)/(dx)=2x^2-3 ?

The solution for 3y(dy)/(dx)=2x^2-3 is y=(sqrt(-18x+4x^3+c_{1)})/3 ,y=-(sqrt(-18x+4x^3+c_{1)})/3

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3y(dy)/(dx)=2x^2-3

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

y = \frac{- 18 x + 4 x ^{3} + c _{1}}{3}, y = - \frac{- 18 x + 4 x ^{3} + c _{1}}{3}

Solution steps

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

Solve separable ODE:

y = \frac{- 18 x + 4 x ^{3} + c _{1}}{3}, y = - \frac{- 18 x + 4 x ^{3} + c _{1}}{3}

y = \frac{- 18 x + 4 x ^{3} + c _{1}}{3}, y = - \frac{- 18 x + 4 x ^{3} + c _{1}}{3}

\frac{d y}{d x} = y^{2} + 2 y - 15

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

y^{^{'}} = e^{t} \cdot y^{2}

y^{^{'}} = 6 y - 5

3 \frac{d x}{d t} = \frac{3 x}{t}

What is the solution for 3y(dy)/(dx)=2x^2-3 ?
The solution for 3y(dy)/(dx)=2x^2-3 is y=(sqrt(-18x+4x^3+c_{1)})/3 ,y=-(sqrt(-18x+4x^3+c_{1)})/3