pk415
Jul15-08, 10:37 PM
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
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.
xU_x + (y^2+1)U_y = U-1; U(x,x) = e^x
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.