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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?

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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.

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    $\begingroup$ I'm looking at the equation you obtained and I don't see how it's separable. We have $$\frac{d}{dt}\left(\ln(1+t^2)\frac{df}{dt}\right)=\ln(1+t^2)\frac{\beta^2}{1+t^2}f$$ how to separate variables here? I'm likely missing something because my basics are rusty $\endgroup$ – Yuriy S Jan 23 '18 at 9:19
  • $\begingroup$ The above equation is "separable" because you can rewrite it in the following form: $$ d\left(\ln(1+t^2)\frac{df}{dt}\right)=\ln(1+t^2)\frac{\beta^2}{1+t^2}f(t)dt $$ So integrating both sides yields: $$ \ln(1+t^2)\frac{df}{dt}+C_1=\int\ln(1+t^2)\frac{\beta^2}{1+t^2}f(t)\,dt $$ Since $\beta>0$ can be small as we like, we can approximate the integrand on the rhs to zero, leaving with a really separable first order ODE: $$ \frac{df}{dt}=\frac{C_1}{\ln(1+t^2)}\quad\to\quad f(t)+C_2 = C_1\int\frac{dt}{\ln(1+t^2)} $$ $\endgroup$ – TheVal Jan 25 '18 at 10:42
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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:

$$\frac{\partial}{\partial z} {_2F_1} (a,b;c;z)=\frac{a b}{c} {_2F_1} (a+1,b+1;c+1;z)$$

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