What is the solution for xy^'+y=6xy^2,y(1)=-8 ?

The solution for xy^'+y=6xy^2,y(1)=-8 is y=-8/(x(48ln(x)+1))

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xy^'+y=6xy^2,y(1)=-8

x y^{^{'}} + y = 6 x y^{2}, y (1) = - 8

y = - \frac{8}{x ( 48 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = 6 x y^{2}

Solve Bernoulli ODE:

y = - \frac{8}{x ( 48 ln ( x ) + 1 )}

y = - \frac{8}{x ( 48 ln ( x ) + 1 )}

\int_{2}^{13} \frac{1}{x ^{3} - 8} d x

\int (3 x + 4)^{\frac{1}{2}} d x

(x y^{2} + 3 y^{2}) d y - 2 x d x = 0

\frac{d}{d x} (\frac{5}{4 + x})

(4, 1), (7, 10)

What is the solution for xy^'+y=6xy^2,y(1)=-8 ?
The solution for xy^'+y=6xy^2,y(1)=-8 is y=-8/(x(48ln(x)+1))