An ODE and a constraint - is there any hope in solving this? 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?
 A: 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}
