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    Mathematics
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    Calculus A Complete Course
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    Exam 19: Ordinary Differential Equations
  5. Question
    Solve the Initial-Value Problem + 2 + 2y
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Solve the Initial-Value Problem + 2 + 2y

Question 87

Question 87

Multiple Choice

Solve the initial-value problem Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t + 2 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t + 2y = 0; y(0) = 4, Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t (0) = -2.


A) y(t) = 4 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t cos t - 6 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t sin t
B) y(t) = 4 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t cos t - 2 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t sin t
C) y(t) = 4 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t cos t + 2 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t sin t
D) y(t) = 4 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t cos t + 6 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t sin t
E) y(t) = 4 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t cos t + 2 Solve the initial-value problem   + 2   + 2y = 0; y(0)  = 4,   (0)  = -2. A)  y(t)  = 4   cos t - 6   sin t B)  y(t)  = 4   cos t - 2   sin t C)  y(t)  = 4   cos t + 2   sin t D)  y(t)  = 4   cos t + 6   sin t E)  y(t)  = 4   cos t + 2   sin t sin t

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