- #1

- 109

- 2

$$ xy' - y = xy^2 $$

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

$$ e^{∫(-1/x)dx} = 1/x $$

$$ (1/x)[y' + (-1/x)y = y^2] $$

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

$$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$

$$ (1/x)y = (y^2)ln|x| + C $$

$$ y = (xy^2)ln|x|+Cx $$

Wolfram shows me another solution...

$$y = \frac{-2x}{C+x^2}$$

Are both correct?