# 6 - Solution set for Math 155 of Assignment 6 Prob 1 t x(t...

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Solution set for Math 155, of Assignment 6 Prob. 1 x ( t ) = x ( 3 ) + Z t 3 cos ( U - 3 ) du = 1 + sin ( U - 3 ) | t 3 = 1 + sin ( t - 3 ) Prob. 2 h ( t ) = h ( 3 ) + Z t 3 ( 5 - 16 t 2 ) du = - 1 + 5 U - 16 3 U 3 ± ± ± t 3 = - 6 1 + 5 t - 16 3 t 3 + 6 1 - 15 + 16 · 3 3 3 = 129 + 5 t - 16 3 t 3 Prob. 3 p ( 10 ) = p ( 0 ) + Z 10 0 ( 3 U + 1 ) du = 3 U 2 2 + U ± ± ± 10 0 = 150 + 10 = 160 Prob. 4 Z dN 5 - 1 2 N = Z dt, so integrating both sides we have: ln | 5 - 1 2 N | - 1 2 = t + c 1 ln | 5 - 1 2 N | = - t 2 + c 2 | 5 - 1 2 N | = e - t / 2 + c 2 5 - 1 2 N = ± c t / 2 + c 2 5 - 1 2 N = c 3 e - t / 2 - 1 2 N = - 5 + c 3 e - t / 2 N = 10 + ce - t / 2 using the fact that N ( 2 ) = 3 , we have 3 = 10 + c · e - 1 - 7 = c / e c = - 7 e so the ﬁnal solution is N ( t ) = 10 - 7 e · e - t / 2 or N ( t ) = 10 - 7 e 1 - t / 2 1

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Prob. 5 Reading the equation of the line from the semilog plot, we have ln p = 1 - 0 . 43 t so derivating both sides we get dp dt p = - 0 . 43 dp dt = - 0 . 43 p . For the initial condition we observe that for time t
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## This note was uploaded on 06/27/2009 for the course MATH 155 taught by Professor Alanm during the Spring '09 term at Simon Fraser.

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6 - Solution set for Math 155 of Assignment 6 Prob 1 t x(t...

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