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Diffrential equations

  1. Apr 5, 2008 #1
    solve the diffrential equation:
    xy' = x(e^(-y/x)) + y


    answer:
    y = xln|lncx|


    problem:
    i don't get it

    can anyone explain
     
  2. jcsd
  3. Apr 5, 2008 #2

    Hootenanny

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    Try re-writing it as

    [tex]y^\prime = e^{-y/x} + \frac{y}{x}[/tex]

    Does a substitution now spring to mind?
     
  4. Apr 5, 2008 #3
    yeah i get that... but then i don't know what to do... do we substitute y=vx???
     
  5. Apr 5, 2008 #4

    Hootenanny

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    That would indeed be my next step :smile:
     
  6. Apr 5, 2008 #5
    lnC + lnx is lnCx yeah?
     
  7. Apr 5, 2008 #6

    Hootenanny

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    It is indeed. In general ln(a)+ln(b)=ln(ab).
     
  8. Apr 5, 2008 #7
    ok
    no matter what i do i'm not getting that answer! and its not supposed to a hard question!!

    i get this:

    y = x lnCx
     
  9. Apr 5, 2008 #8

    Hootenanny

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    What do you get after the substitution?
     
  10. Apr 5, 2008 #9
    What are the last few steps leading up to your answer? Without seeing your work, it is difficult to say where your error is. Though I am inclined to say that you are probably not taking the natural log of the entire right hand side. That is usually where my error is.

    Edit: I notice Hootenany has got you covered. You are in good hands. Back to my DE!
     
  11. Apr 5, 2008 #10
    i didn't want to write it out at first because 1. i can't write it the fancy form and 2. i'm affraid i made a very stupid embarrasing mistake, but ok this is what i did:

    y' = (e^(-y/x)) + y/x

    then y=vx and so v=y/x


    d(vx)/dx = (e^(-v)) + v

    then i integrate:

    vx = x(e^(-v)) + vx + c

    then i substituted y back in

    is this right so far?
     
  12. Apr 5, 2008 #11

    Hootenanny

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    You need to be careful here since v is not a constant, instead it is a function of x, i.e. v=v(x). Therefore, by the product rule,

    [tex]\frac{d}{dx}vx = x\frac{dv}{dx} + v[/tex]

    Hence, the second line of your working is actually,

    [tex]x\frac{dv}{dx} + v = e^{-v} + v[/tex]

    Edit: Just to add a little more detail, since v is a function of x, in general,

    [tex]\int v(x)dx \neq v\cdot x + c[/tex]
     
    Last edited: Apr 5, 2008
  13. Apr 5, 2008 #12
    yes yes yes! i see

    cheers!
     
  14. Apr 5, 2008 #13

    Hootenanny

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    No problem :smile:
     
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