# Homework Help: First Order PDE Cauchy problem Using Method of Characteristics

1. Jul 15, 2008

### pk415

1. The problem statement, all variables and given/known data
Ok, so I can get through most of this but I can't seem to get the last part... Here is the problem

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

2. Relevant equations

3. 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.

2. Jul 16, 2008

### foxjwill

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

$$\begin{cases} x=C_1 e^{\tan^{-1}y}\\ U=C_2 x+1\\ U=C_3 e^{\tan^{-1}y}+1 \end{cases}$$

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/

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

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