question about a PDE

[forty two]

I'm trying to teach myself PDEs over the summer, so I can take more advanced math classes in the fall. I just started a few days ago, and I'm stuck on a problem. I'm sure the problem probably seems easy, but I can't seem to solve it properly.

(y*u_{x} + x*u_{y} = 0)
(U(0,y) = e^{-{y^{2}}})

I did the following:
(\frac{dy}{dx} = \frac{x}{y}) which after integrating becomes:
(y^{2} = x^{2} + c)

Solving for c, and plugging in the specified condition (U(0,y) = e^{-{y^{2}}}) yields
(y^{2} - x^{2} = c)
(f(c) = e^{-{y^{2}}})
(f(y^{2} - x^{2}) = e^{-{y^{2}}})
(f(y^{2} - 0^{2}) = e^{-{y^{2}}})
(f(y^{2}) = e^{-{y^{2}}})

Now this is where I'm getting a bit confused.
(y^{2} = x^{2} - c) therefore
(f(y^{2}) = e^{-{y^{2}}} = e^{-{x^{2} - c}})

But the correct answer is
(u(x,y) = e^{x^{2}-y^{2}})

I'm pretty sure I've done something wrong and it seems something simple. Any help will be greatly appreciated.

 

[MichaelBiro]

your main issue is that you're implicitly assuming the boundary condition holds everywhere. it doesn't.

i.e. ($u(x,y) = f(c) = f(y^2-x^2) \neq e^{-y^2}$)

 

[forty two]

Duh! I feel dumb. I overlooked the fact that the solution is uniquely determined in (xy) plane only where (x^{2}\leq y^{2})

Thank you

 

[MLBrandow]

forty two,

You wouldn't happen to be in my PDEs class would you? I literally just did this problem 40 minutes ago for homework.

 

[Sanket Patel]

MLBrandow,

I take it you're in Dr. Ewalds PDE class? I'm in the class too. Yea, its funny that someone posted that question, I just finished the homework a while ago also. But I'd be surprised if there is a 3rd FSU student on Quantnet.