271_pdfsam_math 54 differential equation solutions odd

271_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exercises 5.2 19. From the first equation, we conclude that y = x − 4x. Substitution into the second equation yields (x − 4x) = −2x + (x − 4x) ⇒ x − 5x + 6x = 0. The characteristic equation, r 2 − 5r + 6 = 0, has roots r = 2, 3, and so a general solution is x(t) = c1 e2t + c2 e3t ⇒ y (t) = c1 e2t + c2 e3t − 4 c1 e2t + c2 e3t = −2c1 e2t − c2 e3t . We find constants c1 and c2 from the initial condition. 1 = x(0) = c1 e2(0) + c2 e3(0) = c1 + c2 , 0 = y (0) = −2c1 e 2(0) − c2 e 3(0) = −2c1 − c2 ⇒ c1 = −1, c2 = 2 . Therefore, the answer to this problem is x(t) = −e2t + 2e3t , y (t) = 2e2t − 2e3t . 21. To apply the elimination method, we write the system using operator notation: D 2 [x] − y = 0, −x + D 2 [y ] = 0. (5.5) Eliminating y by applying D 2 to the first equation and adding to the second equation gives D 2 D 2 − 1 [x] = 0, which reduces to D 4 − 1 [x] = 0. (5.6) The corresponding auxiliary equation, r 4 − 1 = 0, has roots ±1, ±i. Thus, the general solution to (5.6) is given by x(t) = C1 et + C2 e−t + C3 cos t + C4 sin t. Substituting x(t) into the first equation in (5.5) yields y (t) = x (t) = C1 et + C2 e−t − C3 cos t − C4 sin t. (5.8) 267 (5.7) ...
View Full Document

This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online