How Can Characteristics Simplify PDE Solutions?

[c299792458]

Homework Statement

For the system $\psi_{xx}+y\psi_{yy}+{1\over 2}\psi_y=0$ defined on $y<0$.
Show that $\psi(x,y)=f(x+2\sqrt{-y})+g(x-2\sqrt{-y})$ for any functions $f,g$.

Please help

Homework Equations

See above.

The Attempt at a Solution

I think that the characteristics for the system are $\xi={2\over 3}(-y)^{3\over 2}-x,\,\,\,\,\,\eta={2\over 3}(-y)^{3\over 2}+x$ .

So we can reduce the system to ${\partial^2 \psi\over\partial \xi\partial\eta}=0$It is clear to me that $\psi(x,y)=f(\xi)+g(\eta)$ for any functions $f,g$, but I cannot see how I might get the form required.

Perhaps I have done something wrong?

 

[mighty2000]

Any help would be greatly appreciated.Your approach is correct. To get the required form, we can substitute the expressions for ξ and η into the solution for ψ:

ψ(x,y) = f(x+2√(-y)) + g(x-2√(-y))

= f((2/3)(-y)^(3/2)-x+2√(-y)) + g((2/3)(-y)^(3/2)+x-2√(-y))

= f((2/3)(-y)^(3/2)-x+2√(-y)) + g(-(2/3)(-y)^(3/2)+x+2√(-y))

= f((2/3)(-y)^(3/2)-x+2√(-y)) + g(-(2/3)(-y)^(3/2)+x+2√(-y))

= f(4√(-y)) + g(-4√(-y))

= f(x+2√(-y)) + g(x-2√(-y))

So we can see that the solution satisfies the given form.