What is the solution for y^'=-xy^3 ?

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

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y^'=-xy^3

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

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

Solution steps

y^{'} (x) = - x y^{3}

Solve separable ODE:

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

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

x \to 0 - lim (g (x))

\int x^{3} 2^{x} d x

x \to 0 lim (sin (\frac{6}{x}))

\frac{\partial}{\partial u} (u^{2})

\frac{\partial}{\partial z} (e^{y} + x cos (x z))

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