0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t)
0.2(d^2 y)/(dt^2 ) + 1.2dy/dt + 2y = 5cos4t
0.2m^2 + 1.2m + 2 = 0
m = (-b±√(b^2-4ac))/2a
= (-(1.2)± √(〖(1.2)〗^2-4(0.2)(2)))/(2(0.2))
= (-1.2 ±0.4i)/0.4
= -3 ± i
y_h(t) = e^(-3t) (Acosx+Bsinx)
r (t) = 5cos4t
y_p (t) = pcos4t + qsin4t ------------ (1)
y_p^' (t)= -4psin4t + 4qcos4t---------- (2)
y_p^'' (t) = -16pcos4t – 16qsin4t ---------- (3)
Substitute (1), (2) and (3) into 0.2 (d^2 y)/(dt^2 ) + 1.2 dy/dt + 2y = 5cos4t
0.2(-16pcos4t – 16qsin4t) + 1.2(-4psin4t + 4qcos4t) + 2(pcos4t + qsin4t) = 5cos4t
-3.2pcos4t – 3.2qsin4t – 4.8psin4t + 4.8qcos4t + 2pcos4t + 2qsin4t = 5cos4t
Equate the coefficients:
For cos4t:
-3.2p + 4.8q + 2p = 5
-1.2p + 4.8q = 5 ---------- (1)
For sin4t:
-3.2q – 4.8p + 2q = 0
-1.2q – 4.8p = 0 ----------- (2)
Equation 1 x4
19.2 q – 4.8p = 20---------- (3)
-20.4q = -20
q = 50/51
Substitute q =50/51 into equation (2)
-1.2(50/51) – 4.8p = 0
- 4.8p = 20/17
P = - 25/102
y_p (t)= - 25/102cos4t + 50/51sin4t
y (t) =y_h(t) + y_p(t)
= e^(-3t)(Acos t + Bsin t) - 25/102cos4t + 50/51sin4t
Given y (0) = 0.5
A – 25/102 = 0.5
A = 38/51
y’(t) = e^(-3t )(-Asin t + Bcos t) +(Acos t + Bsin t)(-3e^(-3t)) + 50/51sin4t +200/51cos4t
Given y’ (0) = 0
B – 3A + 200/51 = 0
B – 3(38/51) + 200/51 = 0
B = -86/51
Therefore, the equation of motion of the forced oscillation is y(t) = e^(-3t)(38/51cos t – 86/51sin t) – 25/102cos4t + 50/51sin4t
pls help me to check am i correct or not.thanks
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