linear system of two ODEs Consider the following system of ODE:
$$\begin{cases} \dfrac{df}{dt} = \dfrac{\alpha \beta}{1+t^2}g(t), \\  \\
\dfrac{dg}{dt} = \dfrac{\beta}{\alpha}f(t), \\  \\
t \in [-\alpha, \alpha], \\ \\
f(-\alpha) = g(-\alpha) = 1. \end{cases}$$
With $\alpha \ge 1$ and $\beta > 0$ as small as you like. Can this system be solved explicitly? 
 A: Yes.
Take $g(t)$ from the first equation:
$$
g(t) = \frac{1+t^2}{\alpha\beta}\frac{df}{dt}
$$
and take the derivative of the above relation, obtaining $dg/dt$:
$$
\frac{dg}{dt}=\frac{2t}{\alpha\beta}\frac{df}{dt}+\frac{1+t^2}{\alpha\beta}\frac{d^2f}{dt^2}
$$
Now $dg/dt$ is defined also in the second equation:
$$
\frac{dg}{dt}=\frac{\beta}{\alpha}f=\frac{2t}{\alpha\beta}\frac{df}{dt}+\frac{1+t^2}{\alpha\beta}\frac{d^2f}{dt^2}
$$
and this leaves a second-order equation, that is only in terms of $f(t)$:
$$
\frac{1+t^2}{\alpha\beta}\frac{d^2f}{dt^2}+\frac{2t}{\alpha\beta}\frac{df}{dt}-\frac{\beta}{\alpha}f=0
$$
Rearranging the coefficients gives a second order homogeneous ODE with non-constant coefficients:
$$
\frac{d^2f}{dt^2}+\frac{2t}{1+t^2}\frac{df}{dt}-\frac{\beta^2}{1+t^2}f=0
$$
Now, we can write the first two terms as the derivative of a product (I could have used explicitly the integrating factor, but the term $2t/(1+t^2)$ is a strong indicator that a derivative of $\ln(1+t^2)$ is involved) as follows:
$$
\frac{d^2f}{dt^2}+\frac{2t}{1+t^2}\frac{df}{dt} = \frac{1}{\ln(1+t^2)}\frac{d}{dt}\left(\ln(1+t^2)\frac{df}{dt}\right)
$$
The original system has become a second-order ODE, that in turn has become separable. You can take it from here.
A: Another way:
$$\frac{d^2g}{dt^2}=\frac{\beta}{\alpha} \frac{df}{dt}=\frac{\beta^2}{1+t^2}g(t)$$
We obtain a 2nd order ODE:
$$(1+t^2)\frac{d^2g}{dt^2}-\beta^2g=0$$
$$g(-\alpha)=1$$
$$\frac{dg}{dt}(-\alpha)=\frac{\beta}{\alpha}$$
The general solution is (we can check with Wolfram Alpha ):
$$g(t)=C_1 \cdot {_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-t^2 \right)+ \\ + C_2  \cdot t \cdot  {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-t^2 \right)$$
Now we need to substitute the initial conditions to obtain the explicit formula in terms of Hypergeometric function.
$$C_1 \cdot {_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-\alpha^2 \right)- \\ - C_2  \cdot \alpha \cdot  {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-\alpha^2 \right)=1$$
$$C_1=\frac{1+C_2 \alpha {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-\alpha^2 \right)}{{_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-\alpha^2 \right)}$$
I'll leave the second condition to the OP, I'll just provide the formula for the derivative of Hypergeometric function:
$$\frac{\partial}{\partial z} {_2F_1} (a,b;c;z)=\frac{a b}{c} {_2F_1} (a+1,b+1;c+1;z)$$
