Solution

One of the ways to solve this task is by considering conservation of energy.

Drag force takes energy out of the system, adding the momentum once in a period should compensate the losses.

To calculate the energy losses:
$$Q_{-}=\int_{0}^{T} \mu v(x)\,dx = \int_{a}^{b} \mu\cdot v(t)^{2}\,dt$$

We take that:
$$\frac{d v }{d t}+w_{0}^{2}x+\frac{\mu}{m}v=0$$

And with $x(0)=0:$
$$x(t)=A\cdot e^{-\lambda t /2}sin(wt), \lambda=\frac{\mu}{m}$$

In first let's consider $\frac{2 \pi \lambda}{w}<<1$. Drag is very weak.
That means $w\approx w_{0}$ and also means that we can neglect the exponent while calculating integral, because it will lead only to additional terms of 2nd (and greater, thus very small) power of $\lambda$. (one can check by Taylor series).

SO, $v(t)=v_{0}cos(w_{0}t)$,
$$Q_{-}=\int_{0}^{\frac{2 \pi}{w_{0}}} \lambda m v_{0}^{2} \cdot cos^{2}(w_{0}t)\,dt=\lambda m v_{0}^{2} \pi /w_{0}$$
Where $v_{0}$ is speed right after impulse boost.
$$Q_{+}=\frac{m}{2}(v_{0}^{2}-(v_{0}-\frac{p}{m})^{2})\approx pv_{0},$$since $p$ is small, because of weak drag.

Eventually we have $v_{0}=\frac{p w_{0}}{\lambda m \pi }$.
Also, since it is decaying harmonic oscillator, we could take $v_{0}\cdot e^{-\frac{1}{2}\frac{2 \pi \lambda }{w}}=v_{0}-\frac{p}{m}$. And instantly get $$v_{0}=\frac{p}{m}\frac{1}{1-exp(-\frac{\pi \lambda}{w})}\approx \frac{p w_{0}}{\lambda m \pi }$$.

Note that we can't use it for the other case, since the solutions $x(t)$ are equal only in complex form, and in real they're two different types of motion.
One has decaying sinusoid, and another have just multiplication of decaying real exponents: $x(t)=e^{-\lambda/2t} \cdot A e^{-\sqrt{\lambda^2/4 - ω_0^2}t}$. So the solution from Savchenko book is not correct.

Physically, solution shows that body will eventually reach $x=0$ with zero velocity. So, $v=\frac{p}{m}$.