What is the solution for y^'=xy^3(1+x^2) ?

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

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y^'=xy^3(1+x^2)

y^{^{'}} = x y^{3} (1 + x^{2})

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

Solution steps

y^{'} (x) = x y^{3} (1 + x^{2})

Solve separable ODE:

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

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

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

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

\frac{d y}{d x} y = e^{x^{2} l n (x)}

y^{^{'}} = - x y + 4 x

\frac{d}{d x} (tan (y)) = x

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