Question

G.3.a.i) Forced by 5 Sin[1.5 t) Here are three plots of solutions of a random forced exponential diffeq r>0) with random starting values on viOl: Bacics Te rRandon[Real, (e.5, 1.] n[1.5 tl Try 1 Try 1 endtime 28; Try 1. Clea starter1 Random[Real, (5, 1e)]; starter2 Randon[Real, (-2, 2}]; starter3 Randon[ Real, -10, -5)] s]; Try 1. Try 1. y1t starter1 E^(-r t) + E^(-r t) Integrate [ E^ (r_ s) f[s], {s, e, t}]; Try 1 Try 1 узit starter3 E(-r t)+ E^(-r t) Integrate[E (r s) f[s], (s, e, tj]; Try 1 {{Thickness [0.018], Blue, (Thickness [e.014], Red}, {Thickness [0.01], orange}, PlotRange endtine), Plotst -> All, AspectRatio -> Automatic, AxesLabel -> {"t", "}, Plotlabel -> "y'[t] + r y[t] = f[t] with r > 8"] lotf4 t], t, Lesson 2: _ y't]ry-f[t] with r> 0 Qu1tle TUD Tu Ty 1 Try Try 1 Try Ty 1. Try Try Throw in this plot of the forcing function ft): Try 1 (t, e, endtime}, PlotRange -> All, Plotstyle -> {{Thickness [e.02], Black)}]; Showl setup, fplot] Tansforn Fourier Ar 2019 LBUC NetMath All trademarks and registered trademarks are the property of their respective owners Ouf y't]ryt] f[t] with r > 0 Throw in this plot of the forcing function ft): In12 fplot Plot[f[t], {t, e, endtime}, PlotRange -> All, PlotStyle -> {{Thickness [0. 02], Black}}]; Show[setup, fplot] Ouf13 y't]ryt] f [t] with r > 0 Using only your eyes, explain in general terms how the solution plots are related to the plot of the forcing function ft].

Answer 1

The forcing function (in black) in essence forces the system to resonate with it (although with a slight delay)

It can be seen that the different solutions start out from different positions (blue, red, orange graphs) but eventually fall in line with the forcing function. Thus we can say that eventually the forcing function forces out system to resonate with it drowning any effects of initial position or velocity.