H.W.1 Solve the initial value problem y" + y' + 0.25y = 0. MO) = 3.0. y'(0) =- -3.5. y = (3 - 21)e-0.5, Answer H.W.2

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Chapter2: Second-order Linear Odes
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10:41
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1 jo 1
Homework
H.W.1
Solve the initial value problem
y" + y' + 0.25y = 0,
y(0) = 3.0, y'(0) = -3.5.
y = (3 - 2r)e-0.5a
Answer
H.W.2
Ex.2
How does the motion in Example I change if we change the damping constant e from one to another of the
following three values, with y0) 0.16 and y'(0) = 0 as before?
(II) e- 60 kg/sec.
H.W.
(II) e= 10 kg/sec.
H.W.
(1) e- 100 kg/sec.
Solution.
H.W.3
Solve the initial value problem
(6)
y" + 3y' + 2.25y = - 10e-15r
y(0) = 1.
y'(0) = 0.
Continue solving and Find the particular
solution
y = (1+ 1.5x)e-1.r - 5xe-1r - (1 + 1.5r - S-1ar
H.W.4
Solve the initial value problem
(7)
y" + 2y' + 0.75y = 2 cos x - 0.25 sin x + 0.09x,
y(0) = 2.78.
y'(0) = -0.43.
Continue solving and Find the particular
solution
y- 3.le + sinx + 0.12r - 0.32.
Transcribed Image Text:10:41 elearning.uowa.edu.iq 1 jo 1 Homework H.W.1 Solve the initial value problem y" + y' + 0.25y = 0, y(0) = 3.0, y'(0) = -3.5. y = (3 - 2r)e-0.5a Answer H.W.2 Ex.2 How does the motion in Example I change if we change the damping constant e from one to another of the following three values, with y0) 0.16 and y'(0) = 0 as before? (II) e- 60 kg/sec. H.W. (II) e= 10 kg/sec. H.W. (1) e- 100 kg/sec. Solution. H.W.3 Solve the initial value problem (6) y" + 3y' + 2.25y = - 10e-15r y(0) = 1. y'(0) = 0. Continue solving and Find the particular solution y = (1+ 1.5x)e-1.r - 5xe-1r - (1 + 1.5r - S-1ar H.W.4 Solve the initial value problem (7) y" + 2y' + 0.75y = 2 cos x - 0.25 sin x + 0.09x, y(0) = 2.78. y'(0) = -0.43. Continue solving and Find the particular solution y- 3.le + sinx + 0.12r - 0.32.
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