This preview shows page 1. Sign up to view the full content.
Unformatted text preview: w (0) = 0 and w (0) = 1, we frst diFerentiate the solution ound above, then plug in our initial conditions. This gives w (0) = c 1 + c 2 = 0 , w (0) = 2 + 2 c 1 + 2 2 c 2 = 1 . Solving this system o equations yields c 1 = 1 / (2 2) and c 2 = 1 / (2 2). Thus w ( t ) = 1 2 2 e (2+ 2) t 1 2 2 e (2 2) t = 2 4 e (2+ 2) t e (2 2) t is the desired solution. 180...
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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details