example 5.2     p.61   プログラム掲載はp.62より

eq = y''[t] + a2 y'[t]  +  a1 y[t]  /. {a2 -> 2, 
            a1 -> 5};

qt = t;
ans = DSolve[eq == 5 qt , y[t], t][[1]]

yt =  y[t]  /.  ans;
vt =  D[yt, t]

ans2 = Solve[{(yt /. t ->  0) == 
              0, (vt /. t -> 0) == 5}, {C[1], 
            C[2]}][[1]]

yt /. ans2
zu1 = 
      Plot[{qt, %}, {t, 0, 5}, GridLines ->  Automatic]

another solving of example 5.2

Simplify[
      DSolve[{eq == 5 qt, y[0] == 0, y'[0] == 5}, 
        y[t], t]]

example 5.3(1)

qt2=1

Simplify[
        DSolve[{eq == 5 qt2, y[0] == 0, y'[0] == 5}, 
          y[t], t]];
yt2 = y[t] /. %

zu2 = 
        Plot[yt2, {t, 0, 5}, GridLines ->  Automatic, 
          PlotStyle ->  Hue[0.9]];

example 5.3 (2)

qt3 = Sin[ω t]  /.  ω ->  (Sqrt[5])

Simplify[
        DSolve[{eq == 5 qt3, y[0] == 0, y'[0] == 5}, 
          y[t], t]];
yt3 = y[t] /. %

zu3 = 
        Plot[yt3, {t, 0, 5}, GridLines ->  Automatic, 
          PlotStyle ->  Hue[0.7]];

summary

Show[zu1,  zu2,  zu3, PlotRange ->  {-1.5, 5}]