Chain rule when taking non-dimensionalising an ODE I have a silly question. So let's say we have:
$$\frac{d^{2}x}{dt^{2}} = kx$$
Now let's say we pick $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$. What I don't understand is, if we plug in $Xx_{c}$ for x, and $Tt_{c}$ for t, how come we get:
$$\frac{x_{c}d^{2}X}{t_{c}^{2}dT^{2}}$$
How come the $t_{c}$ is squared? Can someone do the math and explain it to me?  Our professo r just said. it was. chain rule but I am not sure how

You could write the chain rule down as
$$ \frac{d X}{dT} = \frac{d X}{dx} \frac{d x}{dt} \frac{d t}{dT} $$
and from your relationships $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$ obtain
$$ \frac{d x}{dt} = \frac{d X}{dT} \frac{x_c}{t_c} $$
The calculation for the second derivative is analogous, just take the first derivative of the above, $$ \frac{\bullet}{dt} = \frac{\bullet}{dT} \frac{dT}{dt} = \frac{\bullet}{dT} \frac{1}{t_c}$$ and the $t_c$ term gets squared.
