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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int 6 x^{5} d x

\frac{d}{d x} (\frac{7 x ^{2}}{4 e ^{x} - x})

\frac{\partial}{\partial y} (e^{x y z^{3}} y z^{3})

y = - \frac{1}{4} x + 5

f (x) = 3 cos (x), at x = \frac{π}{4}

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