>...but it is close.

Ok let's look at this $C(t)=C0*(1+a cos t)$. I think you are right that if one does Taylor expansion in a then up to first order $(1/(1+a cos(t)) \simeq 1- a cos(t)$, moreover

$$C´(t)/C(t)= -a sin(t)/(1+a cos(t)) \simeq -\frac{sin(t)(1+a cos(t))-a sin(t)cos(t)}{(1+a cos(t))^2}|_{a=0} a = -a sin(t)$$, so one has up to first order in a the equation:

$$ (a cos(t)-1) v + a sin (t) v' = v'' $$

So if a is small this is almost a simple oscillator equation and maybe by some continuity argument one can show that the solutions of this equation are close to solutions to the oscillator, but then this is rather more close to a simple oscillator then to a solution of the Mathieu equation (but which of course has the simple oscillator for q small as a kind of limit equation). Moreover you can't even really say that this is a oscillator plus a Mathieulike term because in the end the $v'$ term has to be taken into account. So I see closeness to the simple oscillator but not really to the Mathieu equation.

Ok let's look at this $C(t)=C0*(1+a cos t)$. I think you are right that if one does Taylor expansion in a then up to first order $(1/(1+a cos(t)) \simeq 1- a cos(t)$, moreover

$$C´(t)/C(t)= -a sin(t)/(1+a cos(t)) \simeq -\frac{sin(t)(1+a cos(t))-a sin(t)cos(t)}{(1+a cos(t))^2}|_{a=0} a = -a sin(t)$$, so one has up to first order in a the equation:

$$ (a cos(t)-1) v + a sin (t) v' = v'' $$

So if a is small this is almost a simple oscillator equation and maybe by some continuity argument one can show that the solutions of this equation are close to solutions to the oscillator, but then this is rather more close to a simple oscillator then to a solution of the Mathieu equation (but which of course has the simple oscillator for q small as a kind of limit equation). Moreover you can't even really say that this is a oscillator plus a Mathieulike term because in the end the $v'$ term has to be taken into account. So I see closeness to the simple oscillator but not really to the Mathieu equation.