Thread: 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.

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.

and so we actually have or equivalently . That should be easier to integrate.