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}}

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

\frac{d}{d x} (- \frac{6 x}{x ^{2} + 1})

\int 9 - 4 x d x

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

x \to - 1 lim (x^{3} - x^{2} - 5 x + 3)

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))}