What is the solution for yy^'+xy^2=xy^4 ?

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

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yy^'+xy^2=xy^4

y y^{^{'}} + x y^{2} = x y^{4}

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

Solution steps

y y^{'} (x) + x y^{2} = x y^{4}

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}

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

\int x^{5} I n x d x

\frac{d}{d x} (lo g_{10} (\frac{x}{x + 1}))

x \to - 3 lim (f (x))

\int sec^{8} (x) tan (x) d x

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