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

The solution for xy^'+y=-2xy^2,y(1)=7 is y= 7/(x(14ln(x)+1))

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

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

y = \frac{7}{x ( 14 ln ( x ) + 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{7}{x ( 14 ln ( x ) + 1 )}

y = \frac{7}{x ( 14 ln ( x ) + 1 )}

\frac{d}{d x} (lo g_{2} (x - 1) - 2)

y = x^{2} e^{x} - 2 x e^{x} + 6 e^{x}

f (x) = 16 x, at x = 16

(2 x^{- 2} + 3)^{- 3}

\int \frac{1}{1 0 ^{5} x - x ^{2}} d x

What is the solution for xy^'+y=-2xy^2,y(1)=7 ?
The solution for xy^'+y=-2xy^2,y(1)=7 is y= 7/(x(14ln(x)+1))