Suppose $U:\mathbb{R} \to \mathbb{R}$ and $T:\mathbb{R} \to \mathbb{R}$ are functions of $x$ satisfying:
$\begin{cases} U^{2} = \alpha_{1}\ T^{2} - \alpha_{2}\ T - \alpha_{3} \\ \frac{dT}{dx} = \alpha_{4}\ U \ T \end{cases}$
In the above, we have $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} \in \mathbb{R}$.
Is there anyway to solve for $U(x)$ and $T(x)$ given some initial conditions $U(0)=U_{0}$ and $T(0)=T_{0}$?
Since its non-linear, I'm guessing you can't really do anything here. But maybe in some special cases? For example, fixing particular $\alpha_{j}$'s maybe?
Expressing $U$ in terms of $T$, you obtain the first order ODE \begin{equation} \frac{\text{d} T}{\text{d} x} = \pm \alpha_4 T \sqrt{\alpha_1 T^2 - \alpha_2 T - \alpha_3}. \end{equation} Integration yields \begin{equation} x = \int \pm\frac{1}{\alpha_4 T \sqrt{\alpha_1 T^2 - \alpha_2 T - \alpha_3}} \,\text{d} T. \end{equation} You can look up this integral, for example using Wolfram Alpha, which gives you \begin{equation} x = \pm \frac{1}{\alpha_4 \sqrt{-\alpha_3}} \log\left[\frac{-\alpha_2 T-2\alpha_3 + 2\sqrt{-\alpha_3}\sqrt{\alpha_1 T^2-\alpha_2 T - \alpha_3}}{T}\right] + C, \end{equation} where the integration constant can be determined using the initial condition. You can now rewrite this into \begin{equation} T e^{\mp \alpha_4 \sqrt{-\alpha_3}(x-C)} + \alpha_2 T + 2 \alpha_3 = 2\sqrt{-\alpha_3}\sqrt{\alpha_1 T^2 - \alpha_2 T - \alpha_3}, \end{equation} where you can square both sides to obtain a quadratic equation for $T$: \begin{equation} T^2 \left[\left(\alpha_2 + e^{\mp \alpha_4 \sqrt{-\alpha_3}(x-C)} \right)^2 +4 \alpha_3 \alpha_1\right] + T\left[4 \alpha_3 e^{\mp \alpha_4 \sqrt{-\alpha_3}(x-C)}\right] = 0 \end{equation} (as you see, some terms conveniently cancel out). This yields the trivial solution $T = 0$, and the nontrivial solution \begin{equation} T(x) = \frac{-4 \alpha_3 e^{\mp \alpha_4 \sqrt{-\alpha_3}(x-C)}}{\left(\alpha_2 + e^{\mp \alpha_4 \sqrt{-\alpha_3}(x-C)} \right)^2 + 4 \alpha_3 \alpha_1}. \end{equation} So, yes, you can solve this equation explicitly. Keep in mind that since the solution obeys the constraint $U^2 = \alpha_1 T^2 - \alpha_2 T - \alpha_3$, this must in particular hold for the pair of initial conditions $(T(0),U(0))$. In other words, a solution of the above form exists for general $T(0)$, and the other initial condition $U(0)$ has to obey \begin{equation} U(0)^2 = \alpha_1 T(0)^2 - \alpha_2 T(0) - \alpha_3. \end{equation}