What is the solution for y^'+xy=1xy^3 ?

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

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y^'+xy=1xy^3

y^{^{'}} + x y = 1 x y^{3}

y = \frac{1}{e ^{x^{2} - c_{1}} + 1}, y = - \frac{1}{e ^{x^{2} - c_{1}} + 1}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{e ^{x^{2} - c_{1}} + 1}, y = - \frac{1}{e ^{x^{2} - c_{1}} + 1}

y = \frac{1}{e ^{x^{2} - c_{1}} + 1}, y = - \frac{1}{e ^{x^{2} - c_{1}} + 1}

\int_{1}^{6} - x^{2} + 7 x - 6 d x

x \to 2 + lim (\frac{6 x}{x - 2})

\frac{\partial}{\partial x} (ln (1 + x^{2}) - 2 e^{5 y})

\int \frac{1}{x ( ln ( x ) ) ^{p}} d x

\frac{d}{d y} (c y^{- 6})

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