In those considerations that follow, the positive axis is oriented from down-up for forces, velocities and displacements.

Supposing the mass #m# vertically placed over the spring-dashpot, the mass is actuated by

#f-m g = m ddot x#

where #f# is the force exerted by the spring-dashpot.

The spring-dashpot reacts according to

#-f = kx + c dot x#

Joining both equations we have

#m ddot x + c dot x + k x = -mg#

This is a linear non-homogeneous differential equation whose solution is given by

#x = x_h + x_p#

where #x_h# is the homogeneous solution

#m ddot x_h + c dot x_h + k x_h = 0#

and

#x_p# is a particular solution for

#m ddot x_p + c dot x_p + k x_p = -mg#

In this case #x_p = -(mg)/k#

The homogeneous equation has the general solution

#x_h = e^{lambda t}#

Substituting this generic solution we get at

#(m lambda^2+c lambda + k)e^{lambda t} = 0#

The feasible #lambda# are given by

#lambda=(-c pm sqrt(c^2-4mk))/(2m)#. For our case we have

#lambda = (-10 pm sqrt(100-4 xx 25 xx 226))/(50) = -1/5 pm 3i#

so

#x_h = e^{-t/5} e^{pm i 3 t} = e^{-t/5}(cos(3t)pm i sin(3t))#

when #lambda# is a complex conjugate pair, for each pair the solution proposal is as follows.

#x_h = (C_1 cos(3t)+ C_2sin(3t))e^{-t/5}#

(This can be demostrated but requires a lot of algebra.)

The general solution is given by

#x = (C_1 cos(3t)+ C_2sin(3t))e^{-t/5}-(m g)/k# The constants #C_1,C_2# are determined according to the movement initial conditions. So

#{(x(0) = C_1-(m g)/k = 20),
(dot x(0) = -C_1/5+3C_2=41)
:}#

From this system, we obtain the constants

#C_1 = (m g)/k + 20, C_2 = (m g)/(15k) + 15# assuming #g = 10# we have

#C_1 = 21.1062, C_2 = 15.0737#

So finally

#x = (21.1062 cos(3t)+ 15.0737sin(3t))e^{-t/5}-1.10619#