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, $v=\frac{p}{m}$.