I have the following second order DE:
$\xi'' - k^2\xi = 0 $
$\xi(a) = 0$
I've managed to bring it as far as $\xi = -c_2[e^{k(x-2a)}+e^{-kx}]$, by writing the solution as $\xi = c_1e^{kx}+c_2e^{-kx}$, using the given condition and solving for $c_1$.
The problem is, my book gives the solution as $\xi = c\frac{sinh[k(a-x)]}{sinh(ka)}$.
I guess it's not a difficult problem, but I just don't see it. Does anyone have any ideas?
Thanks!
The denominator is just a constant
$$\xi = c\frac{\sinh[k(a-x)]}{\sinh(ka)}=K{\sinh[k(a-x)]}$$
Then mutliply by $e^{-ka}$ and $\sinh(x)=\dfrac {e^x-e^{-x}}2$
$$\xi =\frac R2({e^{[k(a-x)]}-e^{-ka+kx}})e^{-ka}$$
$$\xi =\frac R2({e^{-kx}-e^{-2ka+kx}})$$ Substitute $\frac R2=C$ $$\xi =-C(e^{k(x-2a)}\color{red}{-{e^{-kx}}})$$
Are you sure you don't have a sign mistake somewhere ?