# xy'= yln(xy)

• Mar 27th 2010, 04:32 PM
rebghb
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.
• Mar 27th 2010, 04:34 PM
Drexel28
Quote:

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.

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