I have the differential equation:
$$\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial z^2}$$
to solve it, the author of my book (page 421 Cooney - Biomedical engineering principles) carries out the following replacement:
$$-2\zeta \frac{dX}{d \zeta} = \frac{d^2 X}{d \zeta^2}$$
with $X=1-\frac{C}{C_S}$ and $\zeta =\frac{z}{(4Dt)^{1/2}}$.
How can I come back to the original equation? I tried:
$$\frac{dX}{d\zeta}=\frac{dX}{dC}\frac{dC}{dt}\frac{dt}{d\zeta}$$
$$\frac{dX}{dC}=-\frac{1}{C_S}$$
$$\frac{dt}{d\zeta} = \frac{d}{d\zeta}\left(z^2(\zeta) \frac{1}{4D\zeta^2}\right) = 2z(\zeta)\frac{dz(\zeta)}{d\zeta}\frac{1}{4D\zeta^2}+\frac{-2z^2(\zeta)}{4D\zeta^3}$$
Now I evaluate $dz/d\zeta$:
$$z=\zeta (4Dt)^{1/2}$$
$$\frac{dz}{d\zeta}=(4Dt)^{1/2}$$
Thus:
$$\frac{dt}{d\zeta} = \frac{d}{d\zeta}\left(z^2(\zeta) \frac{1}{4D\zeta^2}\right) = 2z(\zeta)\frac{dz(\zeta)}{d\zeta}\frac{1}{4D\zeta^2}+\frac{-2z^2(\zeta)}{4D\zeta^3}= \frac{2 \zeta (4Dt)^{1/2} (4Dt)^{1/2}}{4D\zeta^2}-\frac{2\zeta^2 4Dt}{4D\zeta^3}= \frac{2t}{\zeta}-\frac{2t}{\zeta}=0$$
What did I do wrong?
Thank you so much for your time.