What is the solution for xy^'+y=-7xy^2,y(1)=-6 ?

The solution for xy^'+y=-7xy^2,y(1)=-6 is y= 6/(x(42ln(x)-1))

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xy^'+y=-7xy^2,y(1)=-6

x y^{^{'}} + y = - 7 x y^{2}, y (1) = - 6

y = \frac{6}{x ( 42 ln ( x ) - 1 )}

Solution steps

x y^{'} (x) + y = - 7 x y^{2}

Solve Bernoulli ODE:

y = \frac{6}{x ( 42 ln ( x ) - 1 )}

y = \frac{6}{x ( 42 ln ( x ) - 1 )}

\int 14 - (20 - 3 16 - 2 x) d x

y^{- 1}

\frac{\partial}{\partial y} (\frac{1}{y ^{2}})

\frac{d y}{d x} = 12 - y

x^{- 3}

What is the solution for xy^'+y=-7xy^2,y(1)=-6 ?
The solution for xy^'+y=-7xy^2,y(1)=-6 is y= 6/(x(42ln(x)-1))