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If $v_t=cv(u-d)$ is scaled to $v_t=v(u-d)$, then what does "scaling $t$ by $c$" apply to?

Particularly, this equation is advanced w.r.t. time $t$. But if one scaled it, then does it mean that e.g.

$\frac{d}{dt}$ is $\frac{d}{d(t/c)}$? Or also $dt$ would be $d(t/c)$?

And if one calculates $v_t$ over some interval, say $t \in [a,b]$, then would this mean that one scales this by taking $t \in [a/c,b/c]$? But the equation does not have $t$ other than as $u(t), v(t)$.

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