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Homework Help: How to solve this ODE?

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]y' = e^{x-y'}[/tex]

    3. The attempt at a solution
    I have no idea how to handle the situation when y' is appeared in the input of a transcendental function. I substituted y'=p to try to find a parameterized solution to this ODE but it leaded me to nowhere.
  2. jcsd
  3. Oct 31, 2011 #2
    That's going to involve the Lambert W function. How about I show you something and then you apply it to your problem:

    If we have the expression:


    Then I can take the W function of both sides:


    and since the W function is the inverse of ae^a, I can write:


    Ok, now can you get your expression into the W-function form and then invert it with the W function to extract the "a" part?
  4. Oct 31, 2011 #3
    but our professor hasn't told us anything about Lambert W function :( I think first I should find some information about the Lambert function and its properties
  5. Nov 2, 2011 #4
    This is what I've done so far:
    y'=p. ex=pep
    -> (dx/dy)ex=(ep+pep)(dp/dy)
    1=(1+p)dp/dy -> dy=(1+p)dp -> y = p + 1/2p2 + C.
    Now If I replace p=y' I'll obtain y = y' + 1/2(y')2 + C. Do I need to solve this ODE to find y? Is what I've done correct so far?
  6. Nov 2, 2011 #5
    That's confussing to read. I belive you have to use the Lambert W-function to solve this. Just divide by [itex]e^{y'}[/itex]:


    see, that's in Lambert W form. Remember any expression of the form ae^a so I can immediately take the W function of both sides just like you would take the inverse sine or any other inverse function of both sides. The property of this inverse reduces the expression then to:


    Now just integrate:

    [tex]\int_{y_0}^y dy=\int_{t_0}^t W(e^x)dx[/tex]

    That's absolutely no difference conceptually than doing the same thing with any other function like sines and cosines so the solution is:

    [tex]y(x)=y_0+\int_{t_0}^{t} W(e^x)dx[/tex]
  7. Nov 2, 2011 #6
    I understand the logic behind your method, That is fine, but I guess the professor wants us to convert the equation into an ODE of the form y'=f(x,p) and then he wants us to find parametric solutions to the ODE. He hasn't told us anything about Lambert's W function yet so I doubt he would accept my solution in your proposed way.
    I want to say that this parametric curve is the solution to the given ODE:
    x=lnp + p (p>0) y= p + 1/2p2 + C.
    Does that make sense?
    Last edited: Nov 2, 2011
  8. Nov 2, 2011 #7
    Ok, I got it. That's perfectly fine. Just didn't understand it. I do now. Thanks.
  9. Nov 3, 2011 #8


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    Homework Helper

    If the professor does not believe in W just tell him you don't believe in e^x. You now have two equations
    y = p + p2/2 + C
    just eliminate p algebraically.
  10. Nov 3, 2011 #9
    . . . equal rights for special functions. End special-function discrimination.:wink:

    He schooled me though. :)
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