First Order PDE Cauchy problem Using Method of Characteristics

[pk415]

Homework Statement

Ok, so I can get through most of this but I can't seem to get the last part... Here is the problem

xU_x + (y^2+1)U_y = U-1; U(x,x) = e^x

Homework Equations

The Attempt at a Solution

Characteristic equations are:

\frac{dx}{x} = \frac{dy}{y^2+1} = \frac{dU}{U-1}

Solving the first and third gives:

\frac{U-1}{x} = c_1

The first and second equation yield:

tan^{-1}(y) - lnx = c_2

Put the two together in the form

c_1 = f(c_2)

\frac{U-1}{x} = f(tan^{-1}(y) - lnx)

Sub in the Cauchy data and you get

\frac{e^x-1}{x} = f(tan^{-1}(x) - lnx)

Now how do I find what my arbitrary function f is? I have spent hours on this. Is there something that relates inverse tan to natural log? Arrggghhhh!

Thanks for any help.

 

[foxjwill]

Where did the function f come from? The way I see it, you end up with three equations:

\begin{cases}<br /> x=C_1 e^{\tan^{-1}y}\\<br /> U=C_2 x+1\\<br /> U=C_3 e^{\tan^{-1}y}+1<br /> \end{cases}

Reconciling those, you get
U=Cxe^{\tan^{-1}y}+1



Oh, and there is a relation between inverse tan and natural log: http://functions.wolfram.com/ElementaryFunctions/ArcTan/02/0001/

\tan^{-1}z={i \over 2}\ln{1-iz \over 1+iz}