m345t2S11Sol - MA 345 Dierential Equations 1(15 points...

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MA 345 Di ff erential Equations Spring 2011 Exam II Solutions Dr. E. Jacobs 1. (15 points) Let A be the following matrix: A = 3 1 1 3 a) Find all eigenvalues of A 0 = det( A - λ I ) = 3 - λ 1 1 3 - λ = (3 - λ ) 2 - 1 so λ 1 = 2 and λ 2 = 4 b) Find the eigenvectors corresponding to the eigenvalues that you found in part (a). We can solve ( A - λ I ) = 0 by matrix reduction: for λ 1 = 2 , 1 1 | 0 1 1 | 0 -→ 1 1 | 0 0 0 | 0 for λ 2 = 4 , - 1 1 | 0 1 - 1 | 0 -→ 1 - 1 | 0 0 0 | 0 Therefore, for λ 1 = 2, the eigenvectors will be any nonzero scalar multiples of 1 - 1 and for λ 2 = 4, the eigenvectors will be any nonzero scalar multiples of ( 1 1 ) c) Find a matrix P so that P - 1 AP is a diagonal matrix. Check your work by multiplying P - 1 AP out. P = 1 1 - 1 1 check: P - 1 AP = 1 2 1 - 1 1 1 3 1 1 3 1 1 - 1 1 = 1 2 2 - 2 4 4 1 1 - 1 1 = 2 0 0 4 2. (10 points) Once again, let A = ( 3 1 1 3 ) . Let X = x ( t ) y ( t ) . Solve the following di ff erential equation: d X dt = AX where X (0) = - 2 6 The general solution can be constructed immediately from the eigenvectors: X = c 1 1 - 1 e 2 t + c 2 1 1 e 4 t The initial condition implies that : X (0) = - 2 6 = c 1 1 - 1 + c 2 1 1 = 1 1 - 1 1 c 1 c 2 = P c 1 c 2 c 1 c 2 = P - 1 - 2 6 = 1 2 1 - 1 1 1 - 2 6 = - 4 2 Therefore, X = - 4 1 - 1 e 2 t + 2 1 1 e 4 t
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3. (10 points) Let y ( t ) denote the position of a mass at the end of a spring relative to its equilibrium position. We can find a formula for y ( t
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