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

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

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y^'+1/x y=-1/x y^2,y(1)=2

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

y = \frac{2}{3 x - 2}

Solution steps

y^{'} (x) + \frac{1}{x} y = - \frac{1}{x} y^{2}

Solve separable ODE:

y = \frac{2}{3 x - 2}

y = \frac{2}{3 x - 2}

\frac{d}{d x} (lo g_{5} (\frac{4}{x ^{2} 1 - x}))

\int - 4 x^{2} e^{- 3 x} d x

x \to 1 lim ((f (x)) x^{2} - 2 x)

h \to 0 lim (h^{2} + 5^{h})

\int e^{- 7 x} d x

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