Methods of Characteristics I have a problem solving the ODE associated with the question, any help will be greatly appreciated.
Use method of characteristics to solve the problem
$(x-y)\dfrac{\partial u}{\partial x}+(x+y)\dfrac{\partial u}{\partial y}=\alpha u$ 
where $\alpha$ is a constant, with initial condition $u(x,0)=x^2,x\geq0$.
By considering your solution explain :
(i) Why initial conditions cannot be specified on the whole $x$-axis
(ii) why a single valued solution is not possible if $\alpha\neq u$
From what I have learnt, I substituted 
$\dfrac{dx}{ds}=x-y$ and $\dfrac{dy}{ds}=x+y$ , but I am not sure how to proceed from here. 

If I well understand, they are two questions :
First : solving the PDE . Second : "By considering your solution explain" the two items (i) and (ii). 
HINT : Solving the PDE with method of characteristics.
$(x-y)\dfrac{\partial u}{\partial x}+(x+y)\dfrac{\partial u}{\partial y}=\alpha u \quad$ where $\alpha$ is a constant.
$$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{du}{\alpha u}$$
First characteristic equation from 
$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{xdx+ydy}{x^2+y^2}=\frac{du}{\alpha u}$
$\frac{1}{2}\ln|x^2+y^2|=\frac{1}{\alpha}\ln|u|+$constant.
$$(x^2+y^2)^{\alpha /2}u=c_1$$
Second characteristic equation from $\frac{dx}{x-y}=\frac{dy}{x+y}$
In order to solve this homogeneous ODE, change of function : $y=xf(x)$
$\frac{dx}{x-xf}=\frac{fdx+xdf}{x+xf}\quad\to\quad \frac{1-f}{1+f^2}f'=x$
$x^2+\ln(f^2+1)-2\tan^{-1}(f)=c_2$
$$x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right)=c_2$$
The general solution of the PDE, expressed on implicit form is :
$$\Phi\left( \left((x^2+y^2)^{\alpha /2}u\right) \: , \: \left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right)\right)  \right)=0$$
where $\Phi(X,Y)$ is any differentiable function of two variables.
Solving the implicit equation for $X$ leads to the explicit equation :
$(x^2+y^2)^{\alpha /2}u=F\left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)$
where $F$ is any differentiable function.
$$u(x,y)=(x^2+y^2)^{-\alpha /2}F\left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)$$
According to the boundary condition : $u(x,0)=x^2=(x^2)^{-\alpha /2}F\left(x^2\right) \quad\to\quad F(x^2)=(x^2)^{1+\frac{\alpha}{2}}$
$$u(x,y)=(x^2+y^2)^{-\alpha /2} \left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)^{1+\frac{\alpha}{2}}$$
