Write the differential equation in the form x(y′/y) = ln x + ln y and let u = ln y.

Then

du/dx = y′/y and the differential equation becomes x(du/dx) = ln x + u or du/dx − u/x = (ln x)/x, which is first-order and linear. An integrating factor is e^{−∫ dx/x} = 1/x, so that (using integration by parts)

d/dx [1/x u] = ln x/x² and u/x = −1/x – (ln x/x) + c.

The solution is

ln y = −1 − ln x + cx or y = e^cx−1/x.