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xyz
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Try this:

dsl = DSolveValue[{0 == -Fmax (1 - Cos[t]) + k x[t] + 
     d Derivative[1][x][t] + m Derivative[2][x][t], x[0] == xStart, 
   x'[0] == vStart}, x[t], t];
dsl1 = dsl /. d -> 2*Sqrt[m*k] + \[Epsilon]ϵ // Simplify;
lim=Limit[dsl1, \[Epsilon]ϵ -> 0]

yielding the following:

(* (1/(k m (k + m)^2))E^(-((k t)/Sqrt[
  k m])) (E^((k t)/Sqrt[k m]) Fmax k^2 m - 3 Fmax k m^2 + 
   2 E^((k t)/Sqrt[k m]) Fmax k m^2 - Fmax m^3 + 
   E^((k t)/Sqrt[k m]) Fmax m^3 - Fmax m^2 Sqrt[k m] t - 
   Fmax (k m)^(3/2) t + k^3 m t vStart + 2 k^2 m^2 t vStart + 
   k m^3 t vStart + k^3 m xStart + 2 k^2 m^2 xStart + k m^3 xStart + 
   k^3 Sqrt[k m] t xStart + 2 k (k m)^(3/2) t xStart + 
   m (k m)^(3/2) t xStart - 
   E^((k t)/Sqrt[k m]) Fmax k (k - m) m Cos[t] - 
   2 E^((k t)/Sqrt[k m]) Fmax (k m)^(3/2) Sin[t])  *)

looking like this

Plot[lim /. {Fmax -> 1, m -> 1, k -> 1, xStart -> 0, vStart -> 1}, {t,
   0, 5}]

enter image description here

Is it, what you are looking for?

Try this:

dsl = DSolveValue[{0 == -Fmax (1 - Cos[t]) + k x[t] + 
     d Derivative[1][x][t] + m Derivative[2][x][t], x[0] == xStart, 
   x'[0] == vStart}, x[t], t];
dsl1 = dsl /. d -> 2*Sqrt[m*k] + \[Epsilon] // Simplify;
lim=Limit[dsl1, \[Epsilon] -> 0]

yielding the following:

(* (1/(k m (k + m)^2))E^(-((k t)/Sqrt[
  k m])) (E^((k t)/Sqrt[k m]) Fmax k^2 m - 3 Fmax k m^2 + 
   2 E^((k t)/Sqrt[k m]) Fmax k m^2 - Fmax m^3 + 
   E^((k t)/Sqrt[k m]) Fmax m^3 - Fmax m^2 Sqrt[k m] t - 
   Fmax (k m)^(3/2) t + k^3 m t vStart + 2 k^2 m^2 t vStart + 
   k m^3 t vStart + k^3 m xStart + 2 k^2 m^2 xStart + k m^3 xStart + 
   k^3 Sqrt[k m] t xStart + 2 k (k m)^(3/2) t xStart + 
   m (k m)^(3/2) t xStart - 
   E^((k t)/Sqrt[k m]) Fmax k (k - m) m Cos[t] - 
   2 E^((k t)/Sqrt[k m]) Fmax (k m)^(3/2) Sin[t])  *)

looking like this

Plot[lim /. {Fmax -> 1, m -> 1, k -> 1, xStart -> 0, vStart -> 1}, {t,
   0, 5}]

enter image description here

Is it, what you are looking for?

Try this:

dsl = DSolveValue[{0 == -Fmax (1 - Cos[t]) + k x[t] + 
     d Derivative[1][x][t] + m Derivative[2][x][t], x[0] == xStart, 
   x'[0] == vStart}, x[t], t];
dsl1 = dsl /. d -> 2*Sqrt[m*k] + ϵ // Simplify;
lim=Limit[dsl1, ϵ -> 0]

yielding the following:

(* (1/(k m (k + m)^2))E^(-((k t)/Sqrt[
  k m])) (E^((k t)/Sqrt[k m]) Fmax k^2 m - 3 Fmax k m^2 + 
   2 E^((k t)/Sqrt[k m]) Fmax k m^2 - Fmax m^3 + 
   E^((k t)/Sqrt[k m]) Fmax m^3 - Fmax m^2 Sqrt[k m] t - 
   Fmax (k m)^(3/2) t + k^3 m t vStart + 2 k^2 m^2 t vStart + 
   k m^3 t vStart + k^3 m xStart + 2 k^2 m^2 xStart + k m^3 xStart + 
   k^3 Sqrt[k m] t xStart + 2 k (k m)^(3/2) t xStart + 
   m (k m)^(3/2) t xStart - 
   E^((k t)/Sqrt[k m]) Fmax k (k - m) m Cos[t] - 
   2 E^((k t)/Sqrt[k m]) Fmax (k m)^(3/2) Sin[t])  *)

looking like this

Plot[lim /. {Fmax -> 1, m -> 1, k -> 1, xStart -> 0, vStart -> 1}, {t,
   0, 5}]

enter image description here

Is it, what you are looking for?

Source Link
Alexei Boulbitch
  • 40k
  • 2
  • 48
  • 100

Try this:

dsl = DSolveValue[{0 == -Fmax (1 - Cos[t]) + k x[t] + 
     d Derivative[1][x][t] + m Derivative[2][x][t], x[0] == xStart, 
   x'[0] == vStart}, x[t], t];
dsl1 = dsl /. d -> 2*Sqrt[m*k] + \[Epsilon] // Simplify;
lim=Limit[dsl1, \[Epsilon] -> 0]

yielding the following:

(* (1/(k m (k + m)^2))E^(-((k t)/Sqrt[
  k m])) (E^((k t)/Sqrt[k m]) Fmax k^2 m - 3 Fmax k m^2 + 
   2 E^((k t)/Sqrt[k m]) Fmax k m^2 - Fmax m^3 + 
   E^((k t)/Sqrt[k m]) Fmax m^3 - Fmax m^2 Sqrt[k m] t - 
   Fmax (k m)^(3/2) t + k^3 m t vStart + 2 k^2 m^2 t vStart + 
   k m^3 t vStart + k^3 m xStart + 2 k^2 m^2 xStart + k m^3 xStart + 
   k^3 Sqrt[k m] t xStart + 2 k (k m)^(3/2) t xStart + 
   m (k m)^(3/2) t xStart - 
   E^((k t)/Sqrt[k m]) Fmax k (k - m) m Cos[t] - 
   2 E^((k t)/Sqrt[k m]) Fmax (k m)^(3/2) Sin[t])  *)

looking like this

Plot[lim /. {Fmax -> 1, m -> 1, k -> 1, xStart -> 0, vStart -> 1}, {t,
   0, 5}]

enter image description here

Is it, what you are looking for?