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{\partial}{\partial x} (e^{y} sin (x y))

\frac{d}{d x} (\frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}})

f (x) = (x - 3) (2 x + 4)

ln (x^{3} - 2 x)

y^{^{'}} = - \frac{1}{40} y

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))