University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Section 16.2 - First-Order Linear Equations - Exercises - Page 16-14: 11



Work Step by Step

The standard form of the given equation is: $s'+\dfrac{4}{(t-1)}s=\dfrac{(t+1)}{(t-1)^3}$ ....(1) The integrating factor is: $e^{\int (4/(t-1)) dx}=(t-1)^4$ In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides. $\int [s(t-1)^4]' dt=\int \dfrac{(t+1)}{(t-1)^3}(t-1)^4 dt$ or, $s(t-1)^4=\dfrac{t^3}{3}-t+c$ Thus, the general solution is as follows: $s=\dfrac{t^3}{3(t-1)^4}-\dfrac{t}{(t-1)^4}+\dfrac{c}{(t-1)^4}$
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