Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 646: 20

Answer

$y = \frac{{\sin t}}{{{t^3}}}$

Work Step by Step

$$\eqalign{ & {t^3}\frac{{dy}}{{dt}} + 3{t^2}y = \cos t,{\text{ }}y\left( \pi \right) = 0 \cr & {\text{Divide both sides by }}{t^3} \cr & \frac{{dy}}{{dt}} + \frac{{3{t^2}y}}{{{t^3}}} = \frac{{\cos t}}{{{t^3}}} \cr & \frac{{dy}}{{dt}} + \frac{3}{t}y = \frac{{\cos t}}{{{t^3}}} \cr & {\text{The differential equation is in the form }}\frac{{dy}}{{dt}} + P\left( t \right)y = Q\left( t \right) \cr & {\text{With }}P\left( t \right) = \frac{3}{t}{\text{ and }}Q\left( t \right) = \frac{{\cos t}}{{{t^3}}} \cr & {\text{Find the integrating factor }}I\left( t \right) = {e^{\int {P\left( t \right)} dt}} \cr & I\left( t \right) = {e^{\int {\frac{3}{t}dt} }} = {e^{3\left| {\ln t} \right|}} = {t^3} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & \left( {\frac{{dy}}{{dt}} + \frac{3}{t}y} \right){t^3} = \left( {\frac{{\cos t}}{{{t^3}}}} \right){t^3} \cr & {t^3}\frac{{dy}}{{dt}} + 3{t^2}y = \cos t \cr & {\text{Write the left side in the form }}\frac{d}{{dt}}\left[ {I\left( t \right)r} \right] \cr & \frac{d}{{dt}}\left[ {{t^3}y} \right] = \cos t \cr & d\left[ {{t^3}y} \right] = \cos tdt \cr & {\text{Integrate both sides}} \cr & {t^3}y = \int {\cos t} dt \cr & {t^3}y = \sin t + C \cr & {\text{Solve for }}y \cr & y = \frac{{\sin t}}{{{t^3}}} + \frac{C}{{{t^3}}},{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}y\left( \pi \right) = 0 \cr & 0 = \frac{{\sin \pi }}{{{\pi ^3}}} + \frac{C}{{{\pi ^3}}} \cr & C = 0 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & y = \frac{{\sin t}}{{{t^3}}} \cr} $$
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