I have a feeling I may be making a trivial mistake here, but I would really appreciate it if someone could verify my method.
I have a 2nd order PDE:
$$u_{xx} - x^2 u_{yy} - \frac{1}{x} u_x$$
I can reduce this to normal form and find the general solution is given by:
$$u(x,y) = f(x^2 - 2y) + g(x^2 + 2y)$$
Now I have to impose the boundary data: $u(1,y) = 1$ and $u_x(1,y) = 2$.
So to solve this I plug in the data, the first condition gives:
$$1 = f(1-2y) + g(1+2y)$$
And the second:
$$2 = 2f'(1-2y) + 2g'(1+2y)$$
Now I differentiate the first equation and simultaneously solve with the second to give:
$f(1-2y) = \frac{1}{2} y + c_1$ and $g(1+2y) = \frac{1}{2}y + c_2$. Now this is where the difficulty is, when i substitute these into the first equation, it is impossible to determine the coefficients.
Thanks for any help
Start with
$$f(1−2y)+g(1+2y)=1 \text{ (1)}$$,
$$f'(1−2y)+g'(1+2y)=1 \text{ (2)}$$.
@Dmoreno's comment reminded us to differentiate (1) against $y$ and obtain
$$0 = f'(1−2y)-g'(1+2y)= \text { (3)}$$.
From (2) and (3) we obtain:
$$f'(1−2y)=g'(1+2y)=1/2 $$ So $$ f(1-2y)=(1-2y)/2+c_1, g(1+2y)=(1+2y)/2+c_2$$
Substituting them into (1) leads to:
$$1=(1-2y)/2+c_1+(1+2y)/2+c_2=1+c_1+c_2 \implies c_2=-c_1$$