newtonian mechanics - Upwards motion time vs Downwards motion time: Are these sources incorrect?

CONTEXT:

When there is no air resistance, the time taken for a particle to go up $T_U$ will of course be equal to the time taken for the same particle to fall down $T_D$. So in this case, we have $$T_U = T_D$$

But when there is a resistive force say $mkv$ or $mkv^2$, where $m$ is the mass of the particle, $k$ is a positive constant and $v$ is the velocity of the particle, we no longer have such nice symmetry properties. Hence $$T_U \neq T_D$$ So it is only natural to wonder which of the two is larger, and to prove it analytically.

THEIR METHOD:

I have found THREE sources that use the same technique to prove that $T_D > T_U$. However, I think the technique used may not be correct.

The technique follows these steps, roughly speaking. I'll omit the details for now, as my question is regarding the validity of the technique, rather than the computations.

1. Form the acceleration-velocity separable differential equation.



2. Find velocity as a function of time using $a=\frac{dv}{dt}$.


3. Re-arrange and make $v$ the subject, then use the fact that $v=\frac{dx}{dt}$ to find a displacement-time equation $x(t)$.


4. Find the time $T_U$ taken to reach maximum height.


5. Find $x(2T_U)$ and observe that $x(2T_U)>0$. Hence, the downwards journey takes longer.

MY ISSUE:

In the above technique, we are using the same $x(t)$ function, which models upwards motion, to model the downwards motion too.

But I do not think this is valid because in the upwards motion, we have two forces working against us: weight force and resistive force. However, when we are moving downwards, the weight force is now assisting the particle's motion. In other words, the $x(t)$ function will no longer be valid if $t>T_U$.

I cannot see any reason why this technique should work, yet I've found three sources all doing the same thing, so I am inclined to believe that I am missing something here, which is why I come to Physics SE. Could somebody shed some light on this scenario?

SOURCE:

I am very uncertain about the line

For $t \geq t_m$, $v<0$ and therefore $f(v)<0$.

because $f(v)=v^2$ gives an easy counter example to this claim.