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How to solve this ODE?

  • Thread starter AdrianZ
  • Start date
  • #1
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Homework Statement


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

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.
 

Answers and Replies

  • #2
1,796
53
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?
 
  • #3
319
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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
 
  • #4
319
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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?
 
  • #5
1,796
53
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]
 
  • #6
319
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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:
  • #7
1,796
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Ok, I got it. That's perfectly fine. Just didn't understand it. I do now. Thanks.
 
  • #8
lurflurf
Homework Helper
2,426
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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.
 
  • #9
1,796
53
If the professor does not believe in W just tell him you don't believe in e^x.
. . . equal rights for special functions. End special-function discrimination.:wink:

He schooled me though. :)
 

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