Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.3 Exercises: 20



Work Step by Step

$y\sqrt{1-x^2}y'-x\sqrt{1-y^2}=0$ $y\sqrt{1-x^2}y'=x\sqrt{1-y^2}$ $\frac{yy'}{\sqrt{1-y^2}}=\frac{x}{\sqrt{1-x^2}}$ $\int \frac{y}{\sqrt{1-y^2}}dy=\int \frac{x}{\sqrt{1-x^2}}dx$ $-\sqrt{1-y^2}=-\sqrt{1-x^2}+C$ $\sqrt{1-y^2}=\sqrt{1-x^2}+C$ $1-y^2=\left(\sqrt{1-x^2}+C\right)^2$ Let's apply our initial condition here. $1-1^2=(\sqrt{1}+C)^2$ $0=(1+C)^2$ $C=-1$ Solving for $y$ and plugging in $C=-1$ we end up with $$y=\sqrt{1-\left(\sqrt{1-x^2}-1\right)^2}$$
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