Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 32



Work Step by Step

By separation of variables, we have $$\frac{dy}{3y-12}=dx$$ Then by integration, we get $$\ln(3y-12)=3x+c\Longrightarrow y=\frac{1}{3}(Ae^{3x}+12).$$ Now, since $y(2)=1$, then $1=\frac{1}{3}(Ae^{6}+12)\Longrightarrow A=-9e^{-6}$. So the general solution is given by $$y=-3e^{3x-6}+4.$$
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