Often in CLRS, when proving recurrences via substitution, Ө(f(n)) is replaced with cf(n).
For example,on page 91, the recurrence
T(n) = 3T(⌊n/4⌋) + Ө(n^2)
is written like so in the proof
T(n) <= 3T(⌊n/4⌋) + cn^2
But can't Ө(n^2) stand for, let's say, cn^2 + n? Would that not make such a proof invalid? Further in the proof, the statement
T(n) <= (3/16)dn^2 + cn^2
<= dn^2
is reached. But if cn^2 +n was used instead, it would instead be the following
T(n)<= (3/16)dn^2 + cn^2 + n
Can it still be proven that T(n) <= dn^2 if this is so? Do such lower order terms not matter in proving recurrences via substitution?
Yes, it does not matter.
T(n) <= (3/16)dn^2 + cn^2 + n still less than or equal to dn^2 if n is big enough. Because as n goes to infinity, two sides of the equation have the same increasing rate (which is n^2), so the lower-order term will never matter if there is a constant number of lower-order terms in the cost function. But if there is not a constant number of them, that is a different story.
Edit: as n goes to infinity, you will find suitable d and c for T(n) <= (3/16)dn^2 + cn^2 + n to be less than or equal to dn^2, for example d = 2 and c = 1