What is the solution for (dy)/(dx)2x^3y+3xy^3=0 ?

The solution for (dy)/(dx)2x^3y+3xy^3=0 is y=-(2x)/(3+2c_{1)x}

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(dy)/(dx)2x^3y+3xy^3=0

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

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

Solution steps

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

Solve separable ODE:

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

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

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

y^{^{'}} = \frac{2}{3 x y + y}

x y \cdot y^{^{'}} = y - 1

x e^{x^{2}} + y y^{^{'}} = 0, y (0) = 1

\frac{d y}{d x} = - \frac{0.8}{cos ( y )}, y (0) = \frac{π}{4}

What is the solution for (dy)/(dx)2x^3y+3xy^3=0 ?
The solution for (dy)/(dx)2x^3y+3xy^3=0 is y=-(2x)/(3+2c_{1)x}