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I'm looking at the following coupled set of differential equations. Because of the symmetry, I'm hoping to be able to write down the solution for $x_n(t)$ and $y_n(t)$ in terms of $f(t)$ and $g(t)$, but I haven't been able to do so. Any suggestions?

\begin{equation} \begin{cases} x_n'(t)=2f'(t)x_{n-1}(t)-g'(t)(1-x_{n-1}(t)-y_{n-1}(t)), \\ y_n'(t)=2g'(t)y_{n-1}(t)-f'(t)(1-x_{n-1}(t)-y_{n-1}(t)), \end{cases} \end{equation} with $x_0(t)=f(t)$, $y_0(t)=(1-f(t))$, $x_n(0)=1$ and $y_n(0)=0$.

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  • $\begingroup$ These aren't really DEs since there are no $x_n$, $y_n$'s on the RHS; after integrating, it's a recursive definition of $(x_n,y_n)$ in terms of the earlier functions. The first step already produces $y_1(x)=2\int_0^x g'(1-f)$, so nothing very explicit can be done here. $\endgroup$
    – Christian Remling
    Commented Jul 30, 2015 at 18:03

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I have a suggestion but it's an approximation method.

Use Euler method to approximate $x_{n+1}(t)-x_{n}(t)\approx x'_n(t)$.

I don't see how you can solve it analytically, beside plugging $x_n(t) = \sum_{k=0}^\infty a_{kn}t^k$ back to the equation and also for $f(t)$ and $g(t)$ using power series and equating coefficients.

These are only suggestions.

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