What is the solution for xy^'+y=-3xy^2,y(1)=-4 ?

The solution for xy^'+y=-3xy^2,y(1)=-4 is y= 4/(x(12ln(x)-1))

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xy^'+y=-3xy^2,y(1)=-4

x y^{^{'}} + y = - 3 x y^{2}, y (1) = - 4

y = \frac{4}{x ( 12 ln ( x ) - 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{4}{x ( 12 ln ( x ) - 1 )}

y = \frac{4}{x ( 12 ln ( x ) - 1 )}

f (x) = ln^{4} (x)

f (x) = \frac{1}{x - 8}

y = 4 - x^{2}

x \to 0 lim (\frac{x}{∣ x ∣} sin (x))

What is the solution for xy^'+y=-3xy^2,y(1)=-4 ?
The solution for xy^'+y=-3xy^2,y(1)=-4 is y= 4/(x(12ln(x)-1))