# Stuck on: PDE with boundry conditions OR series expantion of x^c

1. Nov 19, 2007

### Xbehave

PDE with boundary conditions

Full question
A function u(x,y) has two independent variables x and y and satisfies the 1st order PDE
$$x \frac{du}{dx} - \frac{y}{2}\frac{du}{dy}= 0$$
by first looking for a separable solution u(x,y)=X(x)Y(y), find the general solution of the equation.
determine the u(x,y) that satisfies the boundary condition u(1,y) = 1 + sin y

a. Relevant equations
$$x \frac{du}{dx} = \frac{y}{2} \frac{du}{dy}$$
$$u(x,y)=X(x)Y(y)$$

a. The attempt at a solution
i think ive got this right:
so i sub in X(x)Y(y) and get
$$x Y(y) \frac{dX(x)}{dx} = X(x) \frac{y}{2}\frac{dY(y)}{dy}$$

$$\frac{x}{X(x)}\frac{dX(x)}{dx} = \frac{y}{2Y(y)}\frac{dY(y)}{dy} = c$$
this gives
$$\int\frac{1}{X(x)} dX = \int\frac{c}{x} dx$$

$$ln[\frac{X(x)}{A_{1}}] = c ln[x]$$
$$X = A_{1} x^{c}$$
doing the same for y gives
$$Y = A_{2} y^{2c}$$

so
$$u(x,y) = X(x)Y(y) = A_{1}A_{2}x^{c}y^{2c} = A (x y^{2})^{c}$$

or with $$b=2c$$ $$u=A(y \sqrt{x})^{b}$$

b. relevant equations
however when i apply the boundary condition i get
$$1 + Sin[y] = A y^{b}$$

b. attempts and ideas
i tried a few e^-iy routes e.g $$\frac{e^{-iy}-e^{iy}}{2i} = Sin[y]$$ but i dont think that helps

the only thing i can think is that i have to relate a power series of a^b to 1 + Sin[a] but i cant remember what a^b is is it related to e^b
which is $$\sum\frac{x^{n}}{n!}$$

or have a i gone wrong already?

Last edited: Nov 20, 2007
2. Nov 20, 2007

### Xbehave

is there anyway to Change the title after thinking about it im pretty sure it should just be
"PDE with boundary condition"
but i think im stuck with the current tittle