Losing-mass oscillator - solving using Laplace transform Statement of a problem
Body of mass $M=m+m_0$ is moved out of balance ($x(0)=A$, $\dot{x}(0)=0$) and is subject to restoration force of magnitude $F = -kx$, where $x(t)$ describes position of body in time. As time passes, body loses mass with rate of $v$ $\frac{kg}{s}$, until its mass is equal to $m_0$.
Find $x(t)$.
Attempted solution
Let:

*

*$ω_0^2 = \frac{k}{M}$


*$ω = \frac{v}{M}$
Until $t=t_0=\frac{m}{v}$ position of the body is described by differential equation:
$(1-ft)\ddot{x}=ω^2x$ (1)
for $t>t_0$ body becomes harmonic oscillator with mass $m_0$, therefore solution for $t>t_0$ is given by:
$x_{t>t_{0}}(t)=Acos(\sqrt{\frac{k}{m_0}}t)$
(1) should be solved and $x(t)$ should be continuous at $t=t_0$
After applying Laplace transform and some additional steps I found that $X(s)=\mathcal{L}\{x(t)\}=\frac{A(s-ω)}{(s-ω)^2+ω_{0}^2-ω^2}$
According to this table (points 20. and 22.) it is true that:
$\mathcal{L}\{e^{at}cos(bt)\}=\frac{s-a}{(s-a)^2+b^2}$
and
$\mathcal{L}\{e^{at}cosh(bt)\}=\frac{s-a}{(s-a)^2-b^2}$
I quess that solution involving $cos$ is true when $b^2>0$ and $cosh$ when $b^2<0$. Under such assuption solution for $t>t_0$ is given by:
$x(t)=Ae^{ωt}cos(\sqrt{ω_0^2-ω^2}t)$ when $ω_0^2>ω^2$ (loss of mass is slow)
$x(t)=Ae^{ωt}cosh(\sqrt{ω^2-ω_0^2}t)$ when $ω_0^2\le ω^2$ (loss of mass is fast)

*

*Is this even correct approach?

*If so, how should I proceed?

*If not, how can this problem be solved?

*Correction of other errors would be appreciated.

 A: Suppose we need to solve with the help of Laplace transform.
In the dynamic system
$$
(m(t)+m_0)\ddot x(t)+k x(t) = 0\\
\dot m(t) = v_0
$$
the second DE does not depends on $x(t)$ so it can be solved independently giving $m(t) = v_0 t + C_0$.
Substituting $m(t)$ into the first equation we have
$$
(v_0t+m'_0)\ddot x(t)+k x(t) = 0
$$ 
with $m'_0 = m_0 + C_0$
Applying the Laplace transform we got
$$
v_0\mathcal{L}\left(t \ddot x\right)+m'_0\mathcal{L}\left(\ddot x\right)+k\mathcal{L}\left(x\right) = 0
$$
or
$$
v_0\left(-\frac{d}{ds}\mathcal{L}\left(\ddot x\right)\right)+m'_0\mathcal{L}\left(\ddot x\right)+k\mathcal{L}\left(x\right) = 0
$$
but
$$
\mathcal{L}\left(\ddot x\right) = x^2X(s)-s x_0 - \dot x_0
$$
so
$$
-v_0\left(2sX(s)+s^2 X'(s)+x_0\right)+m'_0\left(s^2X(s)-s x_0-\dot x_0\right)+kX(s) = 0
$$
which is a DE in the complex plane
$$
X'(s) = \frac{1}{s^2 v_0}\left(\left(m'_0s^2-2v_0 s+k\right)X(s)-m'_0v_0x_0 s-v_0 x_0-\dot x_0m'_0 v_0\right)
$$
once solved this differential equation, then we should calculate the Inverse Laplace transform, which is not possible by elementary methods.
The differential equation can be solved without the help of the Laplace Transform giving as result
$$
x(t) = C_1 \sqrt{\frac{k \left(\frac{m_0}{v_0}+t\right)}{v_0}} I_1\left(2 \sqrt{\frac{k \left(\frac{m_0}{v_0}+t\right)}{v_0}}\right)-C_2 \sqrt{\frac{k
   \left(\frac{m_0}{v_0}+t\right)}{v_0}} K_1\left(2 \sqrt{\frac{k \left(\frac{m_0}{v_0}+t\right)}{v_0}}\right)
$$
Here $I_1, K_1$ are the respective Bessel functions.
