## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 73

#### Answer

$$y\left( t \right) = 3\ln \left| t \right| + 6t + 2$$

#### Work Step by Step

\eqalign{ & y'\left( t \right) = \frac{3}{t} + 6 \cr & y\left( t \right) = \int {y'\left( t \right)} dt \cr & then \cr & y\left( t \right) = \int {\left( {\frac{3}{t} + 6} \right)} dt \cr & find{\text{ the general solution}} \cr & y\left( t \right) = \int {\frac{3}{t}} dt + \int 6 dt \cr & y\left( t \right) = 3\int {\frac{1}{t}} dt + 6\int {dt} \cr & y\left( t \right) = 3\ln \left| t \right| + 6t + C \cr & {\text{using the initial condition }}y\left( 1 \right) = 8 \cr & 8 = 3\ln \left| 1 \right| + 6\left( 1 \right) + C \cr & 8 = 6 + C \cr & C = 2 \cr & {\text{the solution to the initial value problem is}} \cr & y\left( t \right) = 3\ln \left| t \right| + 6t + 2 \cr}

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