Question: Given this equation:

$T=\sqrt{(ugx)^2+(T\_0)^2}$

You're asked to isolate $x$. My process was:

$T=ugx + T\_0$ (the square root cancelled the exponents)

$T-T\_0=ugx$

$x=\frac{T-T\_0}{ug}$

But that was the wrong answer, and they instead followed this process:

$T^2=(\sqrt{(ugx)^2+(T\_0)^2})^2$ (squaring both sides)

$T^2-(T\_0)^2=(ugx)^2$

$\pm \sqrt{T^2-(T\_0)^2}=\sqrt{(ugx)^2}$

$\pm \sqrt{T^2-(T\_0)^2}=ugx$ (why no $\pm$ with $ugx$?)

$x=\frac{\pm \sqrt{T^2-(T\_0)^2}}{ug}$

The answer further said that only positive solutions are included, so the final answer is 

$x=\frac{\sqrt{T^2-(T\_0)^2}}{ug}$

Doesn't that line also evaluate to 

$x=\frac{T-T\_0}{ug}$? I'm pretty confused.

Answer: 
Instead of using optimisation, you can just solve a nonlinear equation. For each $\alpha$, evolve $x\_{\alpha},y\_{\alpha}$ by solving the corresponding differential equation (with initial condition $y(0)=\alpha$), and define $F(\alpha) = y\_{\alpha}(1) - g(x\_{\alpha}(1))$. Now you "just" need to solve $F(\alpha)=0$. I believe you can find inspiration in the "shooting method".
I believe that you have no much better choice than trial & error, but this can be recast as a standard equation resolution problem: the unknown is $y(0)$ and the target function is $y(1)-g(x(1))$, where $x(1)$ and $y(1)$ are obtained by numerical integration, starting from $x(0),y(0)$.

You can use a derivative-free method, such as the secant.