Question

How to you find the general solution of `ylnx-xy'=0`?

Answer 1

`y = C_1e^(1/2(log_e x)^2)`

Explanation:

This is a linear homogeneous differential equation. Also is separable. So grouping variables,

`(y')/y=log_e x/x` After integration

`log_e y = 1/2(log_e x)^2+C` so

`y = C_1e^(1/2(log_e x)^2)`