hw7-solutions

hw7-solutions - Mathematics 53 Winter 2011 Solutions to...

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Mathematics 53, Winter 2011 Solutions to Homework 7 Section 3.5. Problem 12: We frst compute the eigenvalues oF the matrix A = b 2 1 2 - 1 2 1 B . The characteristic equation is 0 = det( λI - A ) = b λ - 2 - 1 2 1 2 λ - 1 B = ( λ - 2)( λ - 1) + 1 4 = λ 2 - 3 λ + 9 4 . ThereFore, A has a repeated eigenvalue ( λ = 3 2 ). Consequently, the solution oF the initial value problem is v y ( t ) = (1 - λt ) e λt v y 0 + te λt Av y 0 = p 1 - 3 t 2 P e 3 t 2 b 1 3 B + te 3 t 2 b 2 1 2 - 1 2 1 Bb 1 3 B = p 1 - 3 t 2 P e 3 t 2 b 1 3 B + te 3 t 2 b 7 2 5 2 B = e 3 t 2 b 1 + 2 t 3 - 2 t B . Problem 14: (a) The electric circuit is described by the system oF di±erential equations I ( t ) = 1 L V ( t ) V ( t ) = - 1 C I ( t ) - 1 RC V ( t ) . Here, I ( t ) and V ( t ) denote the current and voltage at time t , respectively. Moreover, R, L, C are given positive constants. We can write this system in matrix notation: d dt b I ( t ) V ( t ) B = b 0 1 L - 1 C - 1 RC I ( t ) V ( t ) B . The eigenvalues oF the coe²cient matrix satisFy 0 = det b λ - 1 L 1 C λ + 1 RC B = λ 2 + 1 RC λ + 1 LC .
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Therefore, λ = - 1 2 RC ± r 1 4 R 2 C 2 - 1 LC .
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hw7-solutions - Mathematics 53 Winter 2011 Solutions to...

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