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Math Help - xy'= yln(xy)

  1. #1
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rebghb View Post
    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.
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