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 504: 10

Answer

$$ y=\frac{1}{2}x^2+\frac{3}{2}.$$ $$ y= 2e^{x-1}.$$

Work Step by Step

For the equation $$\frac{dy}{dx}=x\Longrightarrow y=\frac{1}{2}x^2+c.$$ Using the condition $ y(1)=2$, we obtain that $2=\frac{1}{2}+c $, then $ c=3/2$ and hence the solution is $$ y=\frac{1}{2}x^2+\frac{3}{2}.$$ For the equation $$\frac{dy}{dx}=y\Longrightarrow \frac{dy}{y}=dx \Longrightarrow \ln y =x+c.$$ Using the condition $ y(1)=2$, we obtain that $\ln 2=1+c $, then $ c=-1+\ln 2$ and hence the solution is $$ y=e^{x-1+\ln 2}=2e^{x-1}.$$
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