Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.1 Exercises - Page 403: 29


The particular solution that passes through the point $(0,3)$ is $$y=3e^{-x/4}.$$

Work Step by Step

The solution $y^2=Ce^{-x/2}$ contains an arbitrary constant $C$. We will determine this constant by imposing that the graph of the function passes through the point $(0,3)$ which means that when we put $x=0$ we have to get $y=3$: $$9=Ce^{-0/2} = Ce^0\Rightarrow C=9.$$ Further we have $$y^2=9e^{-x/2}\Rightarrow y=\pm\sqrt{9e^{-x/2}} = \pm3e^{-x/4}.$$ Note that the only allowable solution for $y$ is with the $"+"$ sign because if we take the $"-"$ sign we will get that for $x=0$ $y=-3$, (while by taking $"+"$ we get what is required, i.e. for $x=0$ $y=3$). Finally, the particular solution that passes through the point $(0,3)$ is $$y=3e^{-x/4}.$$
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