Substitute these into the de 2a 2de x dxe 8ax 4b

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: + Dxe +8Ax + 4B + 4De x x 4Dxe +3Ax2 + 3Bx + 3C + 3Dxe x x = x2 + e x . Matching coefficients, we have 3A = 1 8A + 3B = 0 2A + 4 B + 3C = 0 2D = 1 and solving these yields A = 1/3, B = 8/9, C = 26/27, and D = 1/2. Putting everything together, the solution to the given ODE is y = C1 e 3x + C2 e x 1 + x2 3 8 26 1 x+ + xe x . 9 27 2 AMATH 350 3. Page 3 Assignment #4 Solutions- Winter 2013 a) Show that the price satisfies the second-order DE d2 p dp + 8 + 25p = 100. 2 dt dt dq dS dp = = D S with D = 10 8p and = 25p 4. (a) We have dt dt dt first equation and then using the second gives 100. Di erentiating the dp d2 p dp d2dp dDdq dS p = = S8 with D5p + 100 8p and dS = 25p 8 100+Di pe= 100. 2 = 10 or + 25 rentiating the 4. (a) We have 2 = dt = D 2 dtdt dt dt . dt dt dt dt first equation and then using the second gives 00 (b) If the initial value of the price ispp0+and + 25p = 0. derivative of the price 8p0 the initial b) Associated homogeneous equation: 2 8+ 6 2 d2 dp How does d (t . What is8 + 25 0 ) behaviour 2 4 100p price? 3 Auxili...
View Full Document

This document was uploaded on 02/09/2014.

Ask a homework question - tutors are online