Partial D.E. Change of Variables 
Can anyone tell me how to derive the value for alpha and beta, I'm guessing 6 and 1 in one order or another - "via quadratic". In transforming the equation, could someone show also how the operator say for partially differentiating by x is deduced. Thanks.
 A: Start by defining a function $v(\xi,\eta)$ that has the same values as the function $u(x,y)$ when $\xi$ and $\eta$ are related by the change of variables you quoted in the question. In other words,
$$v(\xi, \eta) = u(x,y)$$
when $\xi = x + \alpha\,y$ and $\eta = \beta\,x + y$. Then, by the chain rule,
$$
\frac{\partial u(x,y)}{\partial x} = 
\frac{\partial v(\xi, \eta)}{\partial\xi}\,\frac{\partial\xi(x,y)}{\partial x} +
\frac{\partial v(\xi, \eta)}{\partial\eta}\,\frac{\partial\eta(x,y)}{\partial x} = 
\frac{\partial v}{\partial\xi} +
\beta\,\frac{\partial v}{\partial\eta}
$$
Similarly,
$$
\frac{\partial u(x,y)}{\partial y} = 
\frac{\partial v(\xi, \eta)}{\partial\xi}\,\frac{\partial\xi(x,y)}{\partial y} +
\frac{\partial v(\xi, \eta)}{\partial\eta}\,\frac{\partial\eta(x,y)}{\partial y} = 
\alpha\,\frac{\partial v}{\partial\xi} +
\frac{\partial v}{\partial\eta}
$$
These can be written in a more useful fashion, for the purpose of this question, as follows:
$$\partial_x = \partial_\xi  + \beta\,\partial_\eta
\qquad\mbox{and}\qquad
\partial_y = \alpha\,\partial_\xi + \partial_\eta$$
Next we need the second and mixed partial derivatives:
$$
\partial_{xx} = 
(\partial_\xi  + \beta\,\partial_\eta)
(\partial_\xi  + \beta\,\partial_\eta) =
\partial_{\xi\xi} + 2\,\beta\,\partial_\xi\,\partial_\eta +
\beta^2\partial_{\eta\eta}
$$
$$
\partial_{yy} = 
(\alpha\,\partial_\xi  + \partial_\eta)
(\alpha\,\partial_\xi  + \partial_\eta) =
\alpha^2\partial_{\xi\xi} + 2\,\alpha\,\partial_\xi\,\partial_\eta +
\partial_{\eta\eta}
$$
$$
\partial_{xy} = 
(\partial_\xi  + \beta\,\partial_\eta)
(\alpha\,\partial_\xi  + \partial_\eta) =
\alpha\,\partial_{\xi\xi} + (\alpha\,\beta + 1)\,\partial_\xi\,\partial_\eta + \beta\,\partial_{\eta\eta}
$$
Now we use these in the original differential equation:
$$
2\,\partial_{xx}u + 3\,\partial_{yy}u - 7\,\partial_{xy}u = 
2\,(\partial_{\xi\xi}v + 2\,\beta\,\partial_\xi\,\partial_\eta v +
\beta^2\partial_{\eta\eta}v) + 3\,(\alpha^2\partial_{\xi\xi}v + 2\,\alpha\,\partial_\xi\,\partial_\eta v +
\partial_{\eta\eta}v) - 7\,(\alpha\,\partial_{\xi\xi}v + (\alpha\,\beta + 1)\,\partial_\xi\,\partial_\eta v + \beta\,\partial_{\eta\eta}v) = 0
$$
Collecting together similar terms gives:
$$
(2 + 3\alpha^2 - 7\alpha)\,\partial_{\xi\xi}v +
(2\beta^2 + 3 - 7\beta)\,\partial_{\eta\eta}v +
\big[4\beta + 6\alpha - 7(\alpha\beta + 1)\big]\,\partial_{\xi\eta}v = 0
$$
So, if we choose $\alpha$ and $\beta$ such that
$$
2 + 3\alpha^2 - 7\alpha = 0 \qquad\mbox{and}\qquad 2\beta^2 + 3 - 7\beta = 0
$$
we will have reduced the original differential equation in $u$ to the following differential equation in $v$:
$$\big[4\beta + 6\alpha - 7(\alpha\beta + 1)\big]\,\partial_{\xi\eta}v = 0$$
Provided that the numerical factor multiplying $\partial_{\xi\eta}v$ above isn't zero, we then have to solve the much simpler differential equation:
$$\partial_{\xi\eta}v = 0$$
The solution to this equation is very simple: 
$$v(\xi,\eta) = f(\xi) + g(\eta)$$
where $f$ and $g$ are arbitrary (but differentiable) functions of their respective arguments.
The quadratic equations for $\alpha$ and $\beta$ give the solutions:
$$\alpha = 2 \qquad\mbox{or}\qquad \alpha = 1/3$$
and
$$\beta = 3 \qquad\mbox{or}\qquad \beta = 1/2$$
Only two of the four $(\alpha,\beta)$ combinations result in $4\beta + 6\alpha - 7(\alpha\beta + 1) \ne 0$:
$$\alpha = 2 \qquad\mbox{and}\qquad \beta = 3$$
and
$$\alpha = 1/3 \qquad\mbox{and}\qquad \beta = 1/2$$
So, for each of these combinations you have a solution and, because the equation is linear, any linear combination of them is also a solution. Thus, the most general solution of the original equation is:
$$u(x,y) = f_1(\xi_1) + g_1(\eta_1) + f_2(\xi_2) +g_2(\eta_2)$$
where $\xi_1 = x + 2y$, $\eta_1 = 3x + y$, $\xi_2 = x + y/3$, $\eta_2 = x/2 + y$, and $f_1$, $g_1$, $f_2$, $g_2$ are arbitrary functions.  In other words,
$$u(x,y) = f_1(x + 2y) + g_1(3x + y) + f_2(x + y/3) + g_2(x/2 + y)$$
But, wait. An arbitrary function of $x + y/3$ is also an arbitrary function of $3x + y$. Similarly, an arbitrary function of $x/2 + y$ is also an arbitrary function of $x + 2y$, so we can simplify the solution above to
$$u(x,y) = f(x + 2y) + g(3x + y)$$
with $f$ and $g$ arbitrary. I recommend that you verify, by direct substitution, that the above is a solution of the original equation because: (1) it's generally a good idea, to make sure that no mistakes have been made and (2) I haven't checked and I may have made a mistake somewhere (not intentionally, of course). :)
Now we need to apply the boundary conditions:
$$
u(0,y) = 4y^2 \Rightarrow f(2y) + g(y) = 4y^2
\qquad\mbox{and}\qquad
u(-2y,y) = 0 \Rightarrow f(0) + g(-5y) = 0
$$
The second equation shows that $g(-5y) = -f(0) = \mbox{constant}$, which means that $g(y)$ must be independent of $y$, ie, a constant. Then, the first equation gives us $f(2y) = 4y^2 + \mbox{constant}$, so $f(y) = y^2 + \mbox{constant}$. Then, $u(x,y) = (x+2y)^2 + C$. The boundary conditions then show that $C = 0$. The solution is, finally, simply
$$u(x,y) = (x+2y)^2$$
