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 )}

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

s l o p e (4, 1), (7, 10)

\frac{d}{d x} (\frac{18}{x})

\frac{\partial}{\partial y} (2 cos (x) + cos (x + y))

\int 5 sin^{2} (r) cos (r) d r

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))