What is the solution for xy^'+y=-2xy^2,y(1)=-5 ?

The solution for xy^'+y=-2xy^2,y(1)=-5 is y= 5/(x(10ln(x)-1))

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xy^'+y=-2xy^2,y(1)=-5

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 5

y = \frac{5}{x ( 10 ln ( x ) - 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{5}{x ( 10 ln ( x ) - 1 )}

y = \frac{5}{x ( 10 ln ( x ) - 1 )}

\frac{d}{d x} (tanh (x))

\int \frac{1}{3 - 5 x} d x

7 x - x^{2} (1.6)

y = x^{2} - 1, y = 3

\frac{\partial}{\partial x} (0.4 x - 4)

What is the solution for xy^'+y=-2xy^2,y(1)=-5 ?
The solution for xy^'+y=-2xy^2,y(1)=-5 is y= 5/(x(10ln(x)-1))