Calculus Cal meets resonance Here\'s a plot of the forced undamped oscillator y\

Question

Calculus Cal meets resonance Here's a plot of the forced undamped oscillator y"[t] + 4 y[t] = f[t] with y[0] = 0 and y'[0] = 2 and with f[t] = 5 Sin[t]6: [112]: = endtime = 25 Clear [f, y, t]; f[t_] = 5 Sin [t]6; ndssol = NDSolve [{y"[t] + 4y[t] == f[t], y [0] == 0, y' [0] == 2}, y[t], {t, 0, endtime}]; ndsy[t_] = y [t] /. ndssol [[1]]; ndsplot = Plot [ndsy[t], {t, 0, endtime}, PlotStyle rightarrow {{Red, Thickness [0.01]}}, AxesLabel rightarrow {"t", "y [t]"}] This is showing signs of resonance. When you showed this plot to that dork Calculus Cal, he said, "Look at the characteristic equation for the unforced oscillator y"[t]+4y[t] = 0:" [118]:= Clear [z]; Solve [z2 + 4 = = 0] Highlight all occurrences of the phrase

Explanation / Answer

y''(t)+4y(t)=5sint6

this is the differential equation,we have to solve this,

charactersitics equation,

D2+4 = 0

D = ±i(2)

therefore, complementory equation is

yc(t) = Acos(2t)+Bsin(2t)

Particular integral,

yp(t) = (5sint6)/(D2+4)= (5sint6)/3

therefore, general equation for this differential equation = yc(t)+yp(t) = Acos(2t)+Bsin(2t)+(5sint6)/3

y(t) = Acos(2t)+Bsin(2t)+(5sint6)/3

intial condition for this;

y(0)=0 and y'(0)=2

at t=0;

y(0) = 0=A

y'(t) = -2Asin(2t)+2Bcos(2t)

y'(0)= 2B= 2

hence, B=1/ and A=0

hence, y(t) =(1/)sin(2t)+(5sint6)/3

graph is same as u see in the problem.

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