1. ## xy'= yln(xy)

Hello everyone,

i got this ODE - xy' = yln(xy)
So i did a simple substitution of u for xy, so u' = y + xy'

u' - y = ylnu equivelent to saying u' / 1 + ln u = y
so integrating the RHS would be simple, but i dont know with what i should integrate y, well if i work it out with the differential operators, i guess i must integrate y with dx, but how...

Any help would be very much appreciated.

2. Originally Posted by rebghb
Hello everyone,

i got this ODE - xy' = yln(xy)
So i did a simple substitution of u for xy, so u' = y + xy'

u' - y = ylnu equivelent to saying u' / 1 + ln u = y
so integrating the RHS would be simple, but i dont know with what i should integrate y, well if i work it out with the differential operators, i guess i must integrate y with dx, but how...

Any help would be very much appreciated.
$\displaystyle xy=u\implies y=\frac{u}{x}$ and so we actually have $\displaystyle \frac{u'}{1+\ln(u)}=\frac{u}{x}$ or equivalently $\displaystyle \frac{u'}{u(1+\ln(u))}=\frac{1}{x}$. That should be easier to integrate.

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### xy'=yln(y/x

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