What is the solution for y^'=x^3y^3 ?

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

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y^'=x^3y^3

y^{^{'}} = x^{3} y^{3}

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

Solution steps

y^{'} (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{d x}{d t} = a x

d x - (8 x y + 3 y) d y = 0

(y^{2} + 1) sin (x) d x - y cos (x) d y = 0, y (0) = 0

\frac{d y}{d x} = \frac{y}{2 x + 1}

y^{^{'}} - a^{2} k y + b^{2} = 0

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