Solving ODE with Transcendental Function: e^{x-y'}

[AdrianZ]

Homework Statement

$$y' = e^{x-y'}$$

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.

 

[jackmell]

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:

$$g(x)e^{g(x)}=h(x)$$

Then I can take the W function of both sides:

$$W\left\{g(x)e^{g(x)}\right\}=W(h(x))$$

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

$$g(x)=W(h(x))$$

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?

 

[AdrianZ]

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

 

[AdrianZ]

This is what I've done so far:
y'=p. e^x=pe^p
-> (dx/dy)e^x=(e^p+pe^p)(dp/dy)
1/p.pe^p=(e^p+pe^p)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?

 

[jackmell]

That's confussing to read. I believe you have to use the Lambert W-function to solve this. Just divide by $e^{y'}$:

$$y'e^{y'}=e^x$$

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:

$$y'=W(e^x)$$

Now just integrate:

$$\int_{y_0}^y dy=\int_{t_0}^t W(e^x)dx$$

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

$$y(x)=y_0+\int_{t_0}^{t} W(e^x)dx$$

 

[AdrianZ]

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?

 

[jackmell]

Ok, I got it. That's perfectly fine. Just didn't understand it. I do now. Thanks.

 

[lurflurf]

If the professor does not believe in W just tell him you don't believe in e^x. You now have two equations
p=e^x-p
y = p + p²/2 + C
just eliminate p algebraically.

 

[jackmell]

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.

He schooled me though. :)