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Consider the second-order linear homogeneous ODE $$ y''(t) - a^2y(t) = 0 $$ In the domain $t = [0,T]$. Note that $a > 0$ in all cases. The characteristic polynomial of the equation is $$ (1)r^2 + (0)r + (-a^2) = 0 $$ which has two real roots $$ r_1 = a $$ $$ r_2 = -a $$

The general solution is then $$ y(t) = K_1e^{at} + K_2e^{-at} $$

To solve for $K_1$ and $K_2$, we need initial conditions for both $y(t)$ and $y'(t)$, where $$ y'(t) = aK_1e^{at} - aK_2 e^{-at} $$

Given the initial value $$ y(t=0) = y_0 $$ I find $$ y(t = 0) = y_0 = K_1e^{a(0)} + K_2e^{-a(0)} $$ $$ \rightarrow y_0 = K_1 + K_2 $$

Instead of an initial value for $y'(t=0)$, I instead know there is a zero-gradient condition at the outer boundary, i.e. $$ y'(t=T) = 0 $$ Which leads to the expression $$ y'(t=T)= 0 = aK_1 e^{aT} - aK_2 e^{-aT} $$ $$ \rightarrow K_2 = K_1 e^{2aT} $$ Substituting this back into the expression for $y(0)$ yields $$ y_0 = K_1 + K_1 e^{2aT} $$ $$ \rightarrow K_1 = \frac{y_0}{1 + e^{2aT}} $$ and substituting $K_1$ back into the expression for $K_2$ yields $$ K_2 = \frac{y_0 e^{2aT}}{1 + e^{2aT}} $$

The full solution is then $$ y(t) = \frac{y_0}{1 + e^{2aT}} e^{at} + \frac{y_0 e^{2aT}}{1 + e^{2aT}} e^{-at} $$ $$ \rightarrow y(t) = \frac{y_0}{1 + e^{2aT}} \left(e^{at} + e^{a(2T-t)} \right) $$

Is that right? The solution is consistent with the boundary conditions I specified, but I want to make sure I'm thinking this through.

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    $\begingroup$ It's easy to see it solves the ODE, just rewrite $e^{a(2T-t)}=e^{2aT} e^{-at}$ and note that $e^{2aT}$ does not depend on $t$. Then you just need to double check that it satisfies both of your boundary conditions. $\endgroup$ – Ian Nov 29 '17 at 19:11

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