What is the solution for xy^'+y=-2xy^2,y(1)=-3 ?

The solution for xy^'+y=-2xy^2,y(1)=-3 is y= 3/(x(6ln(x)-1))

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xy^'+y=-2xy^2,y(1)=-3

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 3

y = \frac{3}{x ( 6 ln ( x ) - 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3}{x ( 6 ln ( x ) - 1 )}

y = \frac{3}{x ( 6 ln ( x ) - 1 )}

\frac{d}{d x} (\frac{4}{11 - x})

y = x 2 x + 3

\int 7^{x} tan (7^{x}) d x

x \to \frac{π}{2} lim (e^{s i n (x)})

x \to 0 lim (\frac{x cos ( x )}{x})

What is the solution for xy^'+y=-2xy^2,y(1)=-3 ?
The solution for xy^'+y=-2xy^2,y(1)=-3 is y= 3/(x(6ln(x)-1))