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

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Answer 1 · 2016-11-22T15:48:26

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

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

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

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