hw7-solutions

# B we next consider the case when r 1 c 1 and l 4

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Unformatted text preview: coeﬃcient matrix has a repeated eigenvalue if and only if LC = 4R2 C 2 , or L = 4R2 C . (b) We next consider the case when R = 1, C = 1, and L = 4. In this case, L = 4R2 C , so the coeﬃcient matrix has a repeated eigenvalue (λ = − 1 ). In 2 order to ﬁnd I (t) and V (t), we need to solve the diﬀerential equation λ=− 1 d I (t) 0 4 = −1 −1 dt V (t) with the initial condition I (t) V (t) I (0) 1 = . V (0) 2 Using the solution formula discussed in lecture, we obtain I (t) = eλt (1 − λt) V (t) t t = e− 2 (1 + ) 2 t 1+t = e− 2 . 2 − 2t I (0) I (0) + tA V (0) V (0) 1 1 +t 2 −3 2 t t Therefore, I (t) = e− 2 (1 + t) and V (t) = e− 2 (2 − 2t). Section 4.7. Problem 3: We are going to look for a solution of the form x(t) = u1 (t) x1 (t) + u2 (t) x2 (t) = u1 (t) e−t −1 1 . + u2 (t) et 1 0 In order for the function x(t) to be a solution of the inhomogeneous ODE x′ (t) = −1 −2 e−t , x(t) + t 0 1 it is necessary and suﬃcient that u′1 (t) e−t e−t −1 1 . = + u′2 (t) et t 1 0 2 This gives u′1 (t) = tet + 1 and u′2 (t) = te−t . Therefore,...
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## This note was uploaded on 02/09/2014 for the course MATH 53 taught by Professor Staff during the Winter '08 term at Stanford.

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