Answer
$y = \frac{\sin(t) + C}{t^3} $
Work Step by Step
Recall that $\frac{d}{dt} t^3 = 3t^2$
So $t^3y'+3t^2y=(t^3y)'$
The equation becomes $(t^3y)' = \cos(t)$
$\int(t^3y)' = \int \cos(t)$
$t^3y = \sin (t)+C$
$y = \frac{\sin(t) + C}{t^3} $
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 625: 16
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