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 )}

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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))