# exam3_s - Exam 3 | April 27, 2004 | Solution 1 1. Consider...

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Unformatted text preview: Exam 3 | April 27, 2004 | Solution 1 1. Consider an initial value problem for homogeneous linear system u v ¶ = 0 1 1 0 ¶ u v ¶ , u (0) v (0) ¶ =- 2 2 ¶ . (a) [10 pts] Find the general solution. Eigenvalues of 0 1 1 0 ¶ are r 1 = 1 and r 2 =- 1. Eigenvector for r 1 = 1 is ξ 1 = 1 1 ¶ and eigenvector for r 2 =- 1 is ξ 2 = 1- 1 ¶ . So the general solution is u ( t ) v ( t ) ¶ = c 1 ξ 1 e r 1 t + c 2 ξ 2 e r 2 t = c 1 e t 1 1 ¶ + c 2 e- t 1- 1 ¶ = c 1 e t + c 2 e- t c 1 e t- c 2 e- t ¶ . (b) [5 pts] Solve the initial value problem.- 2 2 ¶ = u (0) v (0) ¶ = c 1 + c 2 c 1- c 2 ¶ ⇒ c 1 = 0 , c 2 =- 2 . ∴ u ( t ) v ( t ) ¶ =- 2 e- t 1- 1 ¶ =- 2 e- t 2 e- t ¶ . (c) [5 pts] Draw the phase portrait corresponding to the solution obtained in (b). That is, plot the parametric curve of the solution u ( t ) v ( t ) ¶ on uv-plane for t ∈ [0 , ∞ ). 0.5 1 1.5 2 2.5 3 v –3 –2 –1 1 2 3 u MATH-2400 | Introduction to Differential Equations | Spring 2004 | Jeong-Rock Yoon Exam 3 | April 27, 2004 | Solution 2 2. Consider a nonhomogeneous linear system u v ¶ = 1 1- 1 1 ¶ u v ¶ +- 1- 1 ¶ . (a) [5 pts] Find the equilibrium solution b Y . b Y =- 1 1- 1 1 ¶- 1- 1- 1 ¶ = 1 2 1- 1 1 1 ¶ 1 1 ¶ = 1 ¶ . (b) [10 pts] Establish the translated homogeneous linear system where the original equilibrium solution b Y is now moved to the origin. Let W = Y- b Y . Then we have W = ( Y- b Y ) = Y = AY + Z = A ( W + b Y ) + Z = AW + ( A b Y + Z ) | {z } =0 = AW. ∴ W = 1 1- 1 1 ¶ W....
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## This homework help was uploaded on 02/24/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.

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exam3_s - Exam 3 | April 27, 2004 | Solution 1 1. Consider...

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