Drag force proportional to $v^2$, physical interpretation Suppose we have an object with mass $m$, initial velocity $u_0$ and the drag force is $-λu^2$.
Solving for $u$ in Newton's 2nd law we get that:
$u(t)= \frac{u_0}{1+ku_0t}$
where $k=\frac{λ}{m}$.
We can clearly see that there was a time $t_0 =-\frac{1}{ku_0}$ (before what we considered as $t=0$) when the velocity was infinite.
I'm not sure what this means. The only way that the object has a finite velocity at a time t is that it had infinite velocity at $t_0$?.
If that's the case, why isn't that a problem for this particular model of drag force?
Also, does this anomaly have to do with the non linearity of the ode (Newton's 2nd law)?
 A: It's basic math: when you're solving a differential equation, you're doing it in a given domain, and the solution is limited to this domain.
In this example, you used $t=0$ as the initial time, so you solved the equation on $\mathbb{R}^+$. You simply cannot extend the solution to negative times, which are outside the resolution domain.
Physically speaking, just because you see velocity decrease doesn't mean it decreased from infinity before you started looking.
A: With the condition that the  velocity is $u_0$ at time $t=0$ you have obtained a relationship between velocity and time, $u(t)= \dfrac{u_0}{1+ku_0t}$.
You could use this relationship to find velocities before time $t=0$ provided that the drag force is still $-\lambda u^2$.
If $t =-\dfrac{1}{ku_0}+\delta t$ where $\delta t \ll \dfrac{1}{ku_0}$ the relationship tells you that $u(-\dfrac{1}{ku_0}+\delta t) \gg u_0$ and that whatever the finite velocity is at $t =-\dfrac{1}{ku_0}$ or before that time, the velocity condition that is imposed, $u(0)=u_0$, could never be satisfied.
A: At t = 0, $ \mu = \mu_0$. At -3 seconds you will have a different initial velocity (which you should know since it occured in the past) in order to solve the equation without v -> infinity. Otherwise you are attempting to extrapolate information.
A: What is the physical meaning?
The physical meaning is, “If you came upon this particle in those circumstances, you know that there must have been some other force acting on it within a finite past time, $\Delta t=(ku_0/m)^{-1}$, or else some reason for the drag force to not have existed before then. If there were no such force setting it into motion, it would have attained a velocity less than $u_0$ before $t=0.$”
In practice this does not form a deep mathematical problem for the physicist. We are not particularly scared of infinities, especially when they happen “off over there.” So if the initial condition that you are specifying is $u_0$ the speed at $t=0$, probably you have chosen $t=0$ because it signifies the time at which some process that was launching this particle at speed $u_0$ has “let go” of it, for example the particle might be a bullet coming out of the barrel of a gun, and that time might be the moment that it leaves the barrel. Off “over there” at negative times we don't really care about what was happening because we are only modeling in this physical scenario how far the bullet goes, how does it slow down in air, all of that stuff.
To give you an extreme example of this, our modern theory of particle physics, the Standard Model, is in a sense known to be inconsistent or incomplete. It has infinities “over there” at very high energies, which are very small scales of wavelength—the theory just breaks down “over there.” But nobody really minds! I mean, there is a whole cottage industry trying to patch those by looking for a deeper understanding of nature, of course... But it remains the central cornerstone of particle physics.
How to elaborate models
More abstractly, in physics we are always trying to model phenomena, and if we ran into this infinity in practice, i.e. “not over there, over here”, then that would mean we need to choose a better model for our system. Possible elaborations if we started predicting infinities for finite measurements:

*

*Maybe the particle has an engine attached to it.

*Maybe the fluid is not in fact at rest in the coordinate frame.

*Maybe the particle has an electric charge and there is an electromagnetic field that we have not measured yet.

*Maybe the particle is spinning in an interesting way that affects how it travels through such a fluid.

*Maybe there's something interesting about vortex shedding off of this particle.

*Maybe the particle starts traveling so slowly that the drag ceases to be quadratic and becomes linear, this tends to happen underneath a certain characteristic force.

So for example we take a skydiver who jumps out of a plane, not only do we not care about the infinity of speed which you have discovered, but in fact we find out that the skydiver’s velocity does not drop to zero... They do not hang in the air at some altitude, they plummet. Obviously the biggest problem is that we have not included gravity. Well, we include gravity and:
$$
m \frac{\mathrm dv}{\mathrm dt} = mg - k v^2,v(0)=0:\\\text{let }\tau = t\sqrt{gk/m}, u =v\sqrt{k/(mg)},\\
u'(\tau)=1-u^2,u(0)=0\\u=\tanh(\tau)$$
By adding the constant force the infinite velocity has in fact disappeared, instead the equations are telling us that if we wanted to continue this solution backwards in time, we need to fire the skydiver upwards so that at $t=0$ the skydiver reaches their maximum height, and then they will fall according to the same equation.
Quadratic drag is just a building block. It is a standard description of “what is the friction due to a fluid if I am less concerned with that fluid’s viscosity and more concerned with just the effort of pushing the mass of the fluid out from in front of me”. The mass flow rate of that fluid will be proportional to $\Delta v$, the relative velocity between the particle and the fluid, as a result that fluid mass will pick up some forward motion that we assume is also proportional to $\Delta v$, total momentum is the product of mass times change in velocity, so the momentum loss must be proportional to $(\Delta v)^2.$ It is a very raw, unvarnished calculation. All of the real details of fluid motion are completely abstracted away, and not in a wondrous way, but rather in a crude way.
