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

The solution for xy^'+y=-2xy^2,y(1)=-2 is y= 2/(x(4ln(x)-1))

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xy^'+y=-2xy^2,y(1)=-2

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

y = \frac{2}{x ( 4 ln ( x ) - 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2}{x ( 4 ln ( x ) - 1 )}

y = \frac{2}{x ( 4 ln ( x ) - 1 )}

f (x) = x^{3} + 4 x^{2} + 5 x, at x = - 1

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

(7, - 5), (- 7, - 5)

\frac{d}{d x} (y (e^{x} - 1))

\int (\frac{x}{5})^{n} d x

What is the solution for xy^'+y=-2xy^2,y(1)=-2 ?
The solution for xy^'+y=-2xy^2,y(1)=-2 is y= 2/(x(4ln(x)-1))