General solution to an PDE/ODE, change of variables OK, this should be a simple homework problem from the text, but I want to be sure I am following the steps through properly because I feel I am missing the very last bit. 
Given: $$\frac{\partial u}{\partial t} - 2\frac{\partial u}{\partial x}=2$$
this should be pretty easy. Let $\alpha = ax + bt$ and $\beta = cx + dt$
by the chain rule: $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial x} + \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial x}= \frac{\partial u}{\partial \alpha}a + \frac{\partial u}{\partial \beta}c$$ and 
$$\frac{\partial u}{\partial t}=\frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial t}= \frac{\partial u}{\partial \alpha}b + \frac{\partial u}{\partial \beta}d$$
plugging this into the original problem I get 
$$\left(b \frac{\partial u}{\partial \alpha}+d\frac{\partial u}{\partial \beta}\right)-2\left(a \frac{\partial u}{\partial \alpha}+c\frac{\partial u}{\partial \beta}\right)=2$$
So far so good. Doing a little simplifying: $\left(b-2a\right) \frac{\partial u}{\partial \alpha}+ \left(-2c+d \right)\frac{\partial u}{\partial \beta}=2$
Now here's where I feel that I am missing something. I know we want to get rid of one of thoe partials to get to a general solution. The simplest thing to do is to pick values for a, b, c, and d that make one of those terms zero, yes? So let's say that b=2 and a=1. That leaves us with $\left(-2c+d \right)\frac{\partial u}{\partial \beta}=2$. 
Moving the terms around, we have $$\frac{\partial u}{\partial \beta}=\frac{2}{\left(-2c+d \right)}$$
Now the $\beta$ is $cx+dt$. And if we integrate the $\frac{\partial u}{\partial \beta}=\frac{2}{\left(-2c+d \right)}$ we should end up with $$u(\beta)=\left(\frac{2}{\left(-2c+d \right)}\right)\beta=\left(\frac{2}{\left(-2c+d \right)}\right)cx+dt= \left(\frac{2cx+2dt}{\left(-2c+d \right)}\right)$$
This is where I feel I am stuck, because ths looks ugly and I sense that I have gone wrong somewhere. I know the answer is that $u=f(x+2t)-x$ and the change of variables values are a=1 b=2 c=1 and d=0 but I feel that I got something wrong here but I wasn't sure. 
Anyhow, any help would be appreciated. Thanks in advance. 
 A: Try using the method of characteristics (see wiki) or have a look at this explanation of the method of characteristics which highlights the method.
A: Your method is in principle right (if not optimal) all the way up to
$$
\frac{\partial u}{\partial\beta} = \frac{2}{d-2c} .
$$
Note that, quite plainly, you're in trouble unless you pick $d-2c\ne0$. Not-so-coincidentally, this boils down to demanding that
$$
\det\left[\begin{array}{cc}a & b \\ c & d\end{array}\right] \ne 0 ;
$$
this is a condition guaranteeing that the new coordinate system $(\alpha,\beta)$ is a coordinate system - i.e., that its coordinate lines intersect transversally. If I were you, I'd select $c$ and $d$ so that $d-2c=1$, but there's a lot of freedom in selecting these parameters. This also means that no one (including your instructor) can speak of 'the change of coordinates.'
Now, integrate the equation above to get
$$
u(\alpha,\beta) = \frac{2}{d-2c}\beta + f(\alpha) .
$$
The arbitraty function $f$ - which you missed above - plays, here, the role of constant of integration. Since you're integrating with respect to $\beta$, this "constant" is $\beta-$independent; it can thus very well depend on $\alpha$. (You have done a similar thing if you have ever calculated potentials corresponding to conservative vector fields in two or more space dimensions.) Substituting for $\alpha$ and $\beta$, one finds
$$
u(x,y) = \frac{2c}{d-2c} x + \frac{2d}{d-2c} t + f(x+2t) .
$$
It might look like this is too arbitrary. It's not: picking different values of $c$ and $d$ will only give you a solution whose first two terms (the one with $x$ and the other with $t$) differ by a function of $x+2t$; this difference can be absorbed into the (anyhow arbitrary) function $f$. That's a fun little thing to prove.
I hope this works; let us know if it didn't.
