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## Main Question or Discussion Point

The diff. eq. F'(t) / [1 - F(t)] = p + q F(t) or F'(t) = [p + q F(t)][1 - F(t)] has the solution F(t) = [1 - exp(-bt)]/[1 + a exp(-bt)] where a = q / p and b = p + q.

How can I write the system of equations with the solution {F1(t), F2(t)} = { [1 - exp(-b*t)]/[1 + a exp(-b*t)] , [1 - exp(-b*(t-T))]/[1 + a2 exp(-b*(t-T))] } in a similar way, with F1' and F2' on the left-hand side, and p and q (or equivalently a and b) on the right-hand side?

I tried defining L1(t) = F1(t)[1 - F2(t - T)], L2(t) = F2(t - T)[1 + F1(t)] and differentiating, but that did not seem to get me to L1' / (1 - L1) = p + q*L1, L2' / (1 - L2) = p + q*L2.

Thanks for any thoughts and ideas.

How can I write the system of equations with the solution {F1(t), F2(t)} = { [1 - exp(-b*t)]/[1 + a exp(-b*t)] , [1 - exp(-b*(t-T))]/[1 + a2 exp(-b*(t-T))] } in a similar way, with F1' and F2' on the left-hand side, and p and q (or equivalently a and b) on the right-hand side?

I tried defining L1(t) = F1(t)[1 - F2(t - T)], L2(t) = F2(t - T)[1 + F1(t)] and differentiating, but that did not seem to get me to L1' / (1 - L1) = p + q*L1, L2' / (1 - L2) = p + q*L2.

Thanks for any thoughts and ideas.

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