- #1
vanceEE
- 109
- 2
$$ xy' = y + xy^2 $$
$$ xy' - y = xy^2 $$
$$ y' + (-1/x)y = y^2 $$
$$ e^{∫(-1/x)dx} = 1/x $$
$$ (1/x)[y' + (-1/x)y = y^2] $$
$$ (1/x)y' - (1/x^2)y = (1/x)y^2 $$
$$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$
$$ (1/x)y = (y^2)ln|x| + C $$
$$ y = (xy^2)ln|x|+Cx $$
Wolfram shows me another solution...
$$y = \frac{-2x}{C+x^2}$$
Are both correct?
$$ xy' - y = xy^2 $$
$$ y' + (-1/x)y = y^2 $$
$$ e^{∫(-1/x)dx} = 1/x $$
$$ (1/x)[y' + (-1/x)y = y^2] $$
$$ (1/x)y' - (1/x^2)y = (1/x)y^2 $$
$$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$
$$ (1/x)y = (y^2)ln|x| + C $$
$$ y = (xy^2)ln|x|+Cx $$
Wolfram shows me another solution...
$$y = \frac{-2x}{C+x^2}$$
Are both correct?