275_pdfsam_math 54 differential equation solutions odd

# 275_pdfsam_math 54 differential equation solutions odd -...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exercises 5.2 Hence, x = −c1 et − 2c2 e2t − c3 e3t . Finally, we ﬁnd z using (5.10). z = − −c1 et − 2c2 e2t − c3 e3t + c1 et + c2 e2t + c3 e3t 27. We eliminate z by expressing z= 1 1 (−x + 4x) = − (D − 4)[x] 4 4 (5.12) = 2c1 et + 4c2 e2t + 4c3 e3t . from the ﬁrst equation and substituting (5.12) into the second and third equations. We obtain 1 2 − (D − 4)[x] + (D − 4)[y ] = 0, 4 1 1 2x + 4y + D − (D − 4)[x] − 4 − (D − 4)[x] 4 4 After some algebra, the above system simpliﬁes to −(D − 4)[x] + 2(D − 4)[y ] = 0, D2 − 8D + 8 [x] − 16y = 0. We use the second equation to ﬁnd that y= Then the ﬁrst equation becomes −(D − 4)[x] + 2(D − 4) ⇒ (D − 4) −1 + 1 D 2 − 8D + 8 [x] = 0 16 [x] = 0 ⇒ (D − 4)D (D − 8)[x] = 0. 1 D 2 − 8D + 8 [x]. 16 (5.13) = 0. 1 D 2 − 8D + 8 8 Solving the characteristic equation, we get r = 0, 4, and 8; so x = c1 e8t + c2 e4t + c3 . 271 ...
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