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

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

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

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

y = - \frac{2}{x ^{4} + c _{1}}, y = - - \frac{2}{x ^{4} + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{x ^{4} + c _{1}}, y = - - \frac{2}{x ^{4} + c _{1}}

y = - \frac{2}{x ^{4} + c _{1}}, y = - - \frac{2}{x ^{4} + c _{1}}

\frac{\partial}{\partial x} (x arcsin (\frac{y}{x}))

(x^{2} + 9)^{2} (x - 9)

(sin (t) cos (t))^{^{'}}

\frac{d ^{5}}{d x ^{5}} (x^{\frac{11}{2}})

x = 8 - y^{2}, x = y^{2}

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