Constants and superposition principle in PDEs Given this partial differential equation: $$ x \frac{\partial u}{\partial x} -\frac 12 y\frac { \partial u}{\partial y} = 0$$
I would like to first find the general solution and then apply the boundary condition: $u(1,y) = 1+ \sin y$ .
I have used a separation of variables technique, and have gotten two ODE's : $$\frac 1X dX = \frac \lambda x dx$$ and $$\frac 1Y dY = \frac{2\lambda}y dy $$
So one solution will be: $u(x,y) = Cx^\lambda y^{2 \lambda} $. I am confused when using the superposition which states that I can sum up all the solutions because lambda can take any value (even a complex number). I don't understand if this means I have to sum over lambda like so: $$u(x,y) = \sum C_{\lambda} x^\lambda y^{2\lambda } $$ or integrate because lambda is a technically a continuous variable: $$u(x,y) = \int_{-\infty}^{+\infty} C_{\lambda} x^\lambda y^{2\lambda } d \lambda$$ I feel like using the sum would be easier to solve the boundary condition problem as I could equate the summand to the summand in the Taylor series for $1+ \sin x $ . If the integral is the right way to proceed, how do I go about evaluating this integral, and how will I be able to solve the BCP from there?
Any help would be great.
 A: $$ x \frac{\partial u}{\partial x} -\frac 12 y\frac { \partial u}{\partial y} = 0$$
In order to make it more understandable we will compare the results of two different approachs.
The method of characteristics leads to this general solution :
$$u(x,y)=F(x\,y^2)$$
$F$ is an arbitrary function.
The method of separation of variables leads to solutions on this forms  with $t=x\, y^{2 }$ :
$$u(x,y) = \sum_{\lambda} C_{\lambda} x^\lambda y^{2\lambda } =\sum_{\lambda} C_{\lambda} t^\lambda$$
$C_\lambda$ are arbitrary constants.
or :
$$u(x,y) = \int C(\lambda) x^\lambda y^{2\lambda } d \lambda=\int C(\lambda) t^\lambda d \lambda$$
$C(\lambda)$ is an arbitrary function.
Since $C_{\lambda}$ are arbitrary constants and since $C(\lambda)$ is arbitrary thus $\sum C_{\lambda} t^\lambda$ or $\int C(\lambda) t^\lambda d \lambda$ are arbitrary functions of $t$ say $F(t)$.
$$u(x,y)=F(t)=F(xy^2)$$
The two methods lead to equivalent forms of the same general solution.
CONDITION : $u(1,y)=1+\sin(y)$
$$u(1,y)=1+\sin(y)=F(y^2)$$
Let $Y=y^2$
$$1+\sin(\sqrt{Y})=F(Y)$$
Now the function $F$ is known. We put it into the above general solution where $Y=xy^2$
$u(x,y)=1+\sin(\sqrt{xy^2})$
$$u(x,y)=1+\sin(y\sqrt{x})$$
