Suppose I throw a ball up in the air (with air friction $b>0$) at time $t=0$ and it lands at time $T_1$. So we have the equation:
$$m\frac{\mathrm{d}v}{\mathrm{d}t} = -bv + mg$$
And suppose I throw a ball up in a vacuum (with the same initial velocity) at time $t=0$ and it lands at time $T_2$. So we have the equation:
$$m\frac{\mathrm{d}v}{\mathrm{d}t} = mg$$
Is $T_1<T_2$?
On
I remember the answer to this because I was a bit surprised when I first did it.
You'll find that the time spent in the air is less than if it were in a vacuum. Let $v_0$ be the initial velocity and $-v_1$ be the final. You can show without so much work that the time taken for a ball to travel up and back down in a vacuum is given by $$\tau=\frac{2v_0}{g}.$$ When the ball travels up, we have $-mg-bv=F$ and we get $$\int^{t_1}_0dt=-\int^0_{v_0}\frac{v}{g+\frac{b}{m}v}dv=\frac{m}{b}\int^{v_0}_0\left(1-\frac{g}{1+\frac{b}{m}v}\right)dv,$$ which means $$t_1=\frac{m}{b}\log\left(1+\frac{bv_0}{mg}\right).$$ It's the same for travelling back down again, and the time $t_2$ required to do this is $$t_2=-\frac{m}{b}\log\left(1-\frac{bv_1}{mg}\right).$$ You get $$t_1+t_2=\frac{m}{b}\log\left(\frac{mg+bv_0}{mg-bv_1}\right)=\frac{v_0+v_1}{g}.$$ Since $v_1$ must be less than $v_0$ you find that $t_1+t_2<\tau$ which means that the answer to your question is yes.
As commented above however, often you have an equation dependent on $v^2.$ In this case, you would find the speed to be given by a formula like $$v(t)=\sqrt{\frac{mg}{b}}\frac{c_1e^{2\sqrt{\frac{bg}{m}t}}-1}{c_1e^{2\sqrt{\frac{bg}{m}t}}+1}=\sqrt{\frac{mg}{b}}\tanh\left(\sqrt{\frac{bg}{m}}t+c_2\right),$$ as opposed to forces of the form $F_0-bv,$ which give formulae like $$v(t)=v_0e^{-\frac{b}{m}t}+\frac{F_0}{b}\left(1-e^{-\frac{b}{m}t}\right).$$
With height as positive in the upward direction, the ODE becomes $\dot{v} = f(v,b) = -{b \over m} v -g$.
Let $t \mapsto v(t,b)$ denote the solution to the ODE for a given $b \ge 0$, and let $t \mapsto x(t,b)$ represent the corresponding height obtained by integrating $v$. Note that the solutions are $C^1$. It is straightforward to write down an explicit solution for both $b=0$ and $b>0$.
The initial conditions are $v(0,b) = v_0 >0$ and $x(0,b) = 0$. We have $v(t,0) = v_0 -gt$, and for $b>0$, $v(t,b) = e^{-{b \over m} t}(v_0 +{mg \over b}) - {mg \over b}$. Integrating gives $x(t,0) = v_0t-{1\over 2} g t^2$, and for $b>0$, $x(t,b) = {m \over b} \left( (1-e^{-{b \over m} t})(v_0 +{mg \over b}) - gt \right)$.
It is straightforward to see that the height is strictly increasing up to some maximum height, and then strictly decreasing after that. In particular, there exist a unique zero crossing for some $t>0$, and that the velocity is strictly negative at this time. Let $T(b)>0$ denote the unique crossing time.
Direct computation gives $T(0) = {2 v_0 \over g}$. We would like to show that $T(b) < T(0)$ for $b >0$.
Suppose $b>0$, then $x(T(b),b) = 0$ gives $ (1-e^{-{b \over m} T(b)})(v_0 +{mg \over b}) = g T(b)$ (with $T(b)>0$, of course). We can rewrite this as $ { v_0 b \over mg } + 1 = { {b \over m} T(b) \over 1-e^{-{b \over m} T(b)} }$.
We will establish the estimate ${x \over 1 - e^{-x}} > 1+{x \over 2}$ holds for $x > 0$. First note that this gives $ { v_0 b \over mg } + 1 > 1+{1 \over 2}{b \over m} T(b)$, from which we obtain $T(b) < {2 v_0 \over g} = T(0)$.
The estimate can be rearranged to ${x-2 \over x+2} > - e^{-x}$, which is clearly true for $x \ge 2$. Now suppose $0<x<2$ and rewrite the inequality as ${2+x \over 2-x} > e^x$. The Taylor series of the left hand side is $-1+\sum_{k=0}^\infty 2 {x^k \over 2^k}$ and the right hand side is $\sum_{k=0}^\infty {x^k \over k!}$. The coefficients of the constant, $x$ and $x^2$ terms are equal, and the left hand coefficient of $x^k$ is ${2 \over 2^k}$ and the right hand coefficient is ${1 \over k!}$ for $k >2$, and since ${2 \over 2^k}>{1 \over k!}$ we see that the inequality holds for $0<x<2$.