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# hw9_s - Homework 9 | Due on April 12(Monday | Solution 1...

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Homework 9 | Due on April 12 (Monday) | Solution 1 1. Let Y := u v and A denote the 2 × 2 matrix for each problem. (a) [10 pt] u v 0 = 3 - 2 2 - 2 u v i) General solution: We can easily calculate two eigenvalues r 1 = 2, r 2 = - 1 and corresponding eigenvectors ξ 1 = 2 1 , ξ 2 = 1 2 . So the general solution is Y = u v = c 1 2 1 e 2 t + c 2 1 2 e - t . ii)–iv) –2 –1 0 1 2 v –2 –1 1 2 u saddle point unstable v) Initial value problem when Y (0) = 2 0 . Since c 1 = 4 3 and c 2 = - 2 3 , the solution is Y = u v = 4 3 2 1 e 2 t - 2 3 1 2 e - t . (b) [10 pt] u v 0 = 1 - 2 3 - 4 u v i) General solution: Eigenvalues r 1 = - 1, r 2 = - 2. Eigenvectors ξ 1 = 1 1 , ξ 2 = 2 3 . So the general solution is Y = u v = c 1 1 1 e - t + c 2 2 3 e - 2 t . ii)–iv) –2 –1 0 1 2 v –2 –1 1 2 u node asymptotically stable v) Initial value problem when Y (0) = 2 0 . Since c 1 = 6 and c 2 = - 2, the solution is Y = u v = 6 1 1 e - t - 2 2 3 e - 2 t . MATH-2400 | Introduction to Differential Equations | Spring 2004 | Jeong-Rock Yoon

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Homework 9 | Due on April 12 (Monday) | Solution 2 (c) [10 pt] u v 0 = 5 3 3 5 u v i) General solution: Eigenvalues r 1 = 2, r 2 = 8. Eigenvectors ξ 1 = - 1 1 , ξ 2 = 1 1 . So the general solution is Y = u v = c 1 - 1 1 e 2 t + c 2 1 1 e 8 t . ii)–iv) –2 –1 0 1 2 v –2 –1 1 2 u node unstable v) Initial value problem when Y (0) = 2 0 . Since c 1 = - 1 and c 2 = 1, the solution is Y = u v = 1 - 1 e 2 t + 1 1 e 8 t . (d) [10 pt] u v 0 = 2 - 5 1 - 2 u v i) Eigenvalues r 1 = i , r 2 = - i . Eigenvectors ξ 1 = 2 + i 1 , ξ 2 = 2 - i 1 . ξ 1 e r 1 t = 2 + i 1 e it = (2 + i )(cos t + i sin t ) cos t + i sin t = (2 cos t - sin t ) + i (cos t + 2 sin t ) cos t + i sin t = 2 cos t - sin t cos t | {z } Re [ ξ 1 e r 1 t ] + i cos t + 2 sin t sin t | {z } Im [ ξ 1 e r 1 t ] So, Y = c 1 Re £ ξ 1 e r 1 t / + c 2 Im £ ξ 1 e r 1 t / = c 1 2 cos t - sin t cos t + c 2 cos t + 2 sin t sin t . ii)–iv) –2 –1 0 1 2 v –2 –1 1 2 u center stable MATH-2400 | Introduction to Differential Equations | Spring 2004 | Jeong-Rock Yoon
Homework 9 | Due on April 12 (Monday) | Solution 3 v) Initial value problem when Y (0) = 2 0 .

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hw9_s - Homework 9 | Due on April 12(Monday | Solution 1...

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