Mathematica code to draw a graph of this differential equation? - wolfram-mathematica

Question

Does anyone know the Mathematica code that will trace the graph below?
Here is the equation for the graph, a second order linear differential equation with constant coefficients:
Here is the graph traced by this equation:
Quote from the book "Times Series Analysis and Forecasting By Example":
... where δ(t ) is an impulse (delta) function that, like a pea shot, at
time t = 0 forces the pendulum away from its equilibrium and a is the
size of the impact by the pea. It is easy to imagine that the curve
traced by this second order differential equation is a damped
sinusoidal function of time although, if the friction or viscosity is
sufficiently large, the (overdamped) pendulum may gradually come to
rest following an exponential curve without ever crossing the
centerline.

Answer 1

eq = m z''[t] + c z'[t] + k z[t] == a DiracDelta[t];
parms = {m -> 1, c -> .1, k -> 1, a -> 1};
sol = First#DSolve[{eq /. parms, z[0] == 1, z'[0] == 0}, z[t], t];
Plot[z[t] /. sol, {t, 0, 70}, PlotRange -> All, Frame -> True,
FrameLabel -> {{z[t], None}, {Row[{t, " (sec)"}], eq}},
GridLines -> Automatic]
Notice that, for zero initial conditions, another option is to use the Control system functions in Mathematica as follows
parms = {m -> 10, c -> 1.2, k -> 4.3, a -> 1};
tf = TransferFunctionModel[a/(m s^2 + c s + k) /. parms, s]
sol = OutputResponse[tf, DiracDelta[t], t];
Plot[sol, {t, 0, 60}, PlotRange -> All, Frame -> True,
FrameLabel -> {{z[t], None}, {Row[{t, " (sec)"}], eq}},
GridLines -> Automatic]
Update
Strictly speaking, the result of DSolve above is not what can be found by hand derivation of this problem. The correct solution should come out as follows
(see this also for reference)
The correct analytical solution is given by
which I derived for this problem and similar cases in here (first chapter).
Using the above solution, the correct response will look like this:
parms = {m -> 1, c -> .1, k -> 1, a -> 1};
w = Sqrt[k/m];
z = c/(2 m w);
wd = w Sqrt[1 - z^2];
analytical =
Exp[-z w t] (u0 Cos[wd t] + (v0 + (u0 z w))/wd Sin[wd t] +
a/(m wd) Sin[wd t]);
analytical /. parms /. {u0 -> 1, v0 -> 0}
(* E^(-0.05 t) (Cos[0.998749 t] + 1.05131 Sin[0.998749 t]) *)
Plotting it:
Plot[analytical /. parms /. {u0 -> 1, v0 -> 0}, {t, 0, 70},
PlotRange -> All, Frame -> True,
FrameLabel -> {{y[t], None}, {Row[{t, " (sec)"}],
"analytical solution"}}, GridLines -> Automatic, ImageSize -> 300]
If you compare the above plot with the first one shown above using DSolve you can see the difference near t=0.