What is the solution for xy^'+y=-8xy^2,y(1)=3 ?

The solution for xy^'+y=-8xy^2,y(1)=3 is y= 3/(x(24ln(x)+1))

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xy^'+y=-8xy^2,y(1)=3

x y^{^{'}} + y = - 8 x y^{2}, y (1) = 3

y = \frac{3}{x ( 24 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = - 8 x y^{2}

Solve Bernoulli ODE:

y = \frac{3}{x ( 24 ln ( x ) + 1 )}

y = \frac{3}{x ( 24 ln ( x ) + 1 )}

\frac{s + 1}{s ^{2} + 4 s + 8}

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What is the solution for xy^'+y=-8xy^2,y(1)=3 ?
The solution for xy^'+y=-8xy^2,y(1)=3 is y= 3/(x(24ln(x)+1))