MathQuest: Differential Equations

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1 MathQuest: Differential Equations Solutions to Linear Systems. Consider the linear system given by dy dt = 4 True or False: Y e t t = is a solution. c False, but I am not very confident Y.. Consider the linear system d Y dt = 4 Y with solution Y e t t = True or False: The function k Y t formed by multiplying Y t by a constant k is also a solution to the linear system. c False, but I am not very confident.. Consider the linear system d Y dt solutions to the linear system. = 4 Y. The functions Y t and Y t are True or False: The function Y t+ Y t formed by adding the two solutions Y t and Y t is also a solution to the linear system. c False, but I am not very confident

2 4. True or False: The functions Y sint t = cost independent. c False, but I am not very confident and Y t = cost sint are linearly 5. True or False: The functions Y t = linearly independent. sint cost and Y t = sint cost are c False, but I am not very confident 6. If we are told that the general solution to the linear homogeneous system Y = AY is Y = c e 4t +c e t, then an equivalent form of the solution is a y = c e 4t +c e 4t and y = c e t +c e t b y = c e 4t +c e t and y = c e 4t +c e t c y = c e 4t +c e 4t and y = c e t +c e t d y = c e 4t +c e t and y = c e 4t +c e t e All of the above f None of the above 7. If Y = e t +e 4t is a solution to the linear homogeneous system Y = AY, which of the following is also a solution? a Y = e t b Y = e t 4e 4t c Y = /4e 4t

3 d All of the above e None of the above 8. The eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system Y = AY are λ = 4 with v =<, > and λ = with v =<, >. What is a form of the solution? a Y = c e 4t +c e t b Y = c e 4t +c e t c Y = c e 4t +c e t d Y = c e 4t +c e t 9. You have a linear, homogeneous, system of differential equations with constant coefficients whose coefficient matrix A has the eigensystem: eigenvalues -5 and - and eigenvectors <, > and < 4,5 >, respectively. Then a general solution to dy dt = A Y is given by: k e a Y = 5t +k e t 4k e 5t +5k e t k e b Y = t 4k e 5t k e t +5k e 5t k e c Y = 5t 4k e t k e 5t +5k e t k e d Y = t +k e 5t 4k e t +5k e 5t. The eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system Y = AY are λ = 4 with v =<, > and λ = with v =<, >. In the long term, phase trajectories: a become parallel to the vector v =<, >. b tend towards positive infinity. c become parallel to the vector v =<, >. d tend towards.

4 e None of the above. If the eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system Y = AY are λ = 4 with v =<, > and λ = with v =<, >, is y a = e 4t a solution? a Yes, it is a solution. b No, it is not a solution because it does not contain λ. c No, it is not a solution because it is a vector. d No, it is not a solution because of a different reason.. The eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system Y = AY are λ = 4 with v =<, > and λ = 4 with v =<, >. What is a form of the solution? a Y = c e 4t b Y = c e 4t c Y = c e 4t d Y = c e 4t +c e 4t +c e 4t +c te 4t +c e 4t. The system of differential equations Y = Y has eigenvalue λ = with multiplicity, and all eigenvectors are multiples of v =<, >. Testing Y = e t, we find that a Y is a solution. b Y is not a solution. 4. The system of differential equations Y = Y has eigenvalue λ = with multiplicity, and all eigenvectors are multiples of v =<, >. One solution to this equation is Y = e t. Testing Y = te t, we find that 4

5 a Y = te t b Y = te t is also a solution. is not a solution. 5. The eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system Y = AY are λ = 4 with multiplicity, and all eigenvectors are multiples of v =<, >. What is the form of the general solution? a Y = c e 4t b Y = c e 4t c Y = c e 4t +c te 4t +c e 4t +c te 4t +e 4t 5

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