1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]g(x)e^{g(x)}=h(x)[/tex]

    Then I can take the W function of both sides:

    [tex]W\left\{g(x)e^{g(x)}\right\}=W(h(x))[/tex]

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

    [tex]g(x)=W(h(x))[/tex]

    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/p.pep=(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]:

    [tex]y'e^{y'}=e^x[/tex]

    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:

    [tex]y'=W(e^x)[/tex]

    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

    lurflurf

    User Avatar
    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
    p=ex-p
    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. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to solve this ODE?
  1. How to solve this ODE? (Replies: 2)

  2. How to solve this ODE? (Replies: 2)

  3. How to solve this ODE? (Replies: 1)

  4. How to solve this ode (Replies: 0)

Loading...