Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.8 Change of Variables - Problems - Page 79: 12

Answer

\[y(x)=x \cos^{-1}\left(\frac{1}{Cx}\right) \]

Work Step by Step

\[\sin\left(\frac{y}{x}\right)(xy'-y)=x\cos\left(\frac{y}{x}\right)\] \[y'=\frac{x\cot\left(\frac{y}{x}\right)+y}{x}\] \[y'=\cot\left(\frac{y}{x}\right)+\frac{y}{x}\;\;\;\ldots (1)\] Substitute $\:y=Vx\;\;\;\ldots (2)$ Differentiate (2) with respect to $x$ \[y'=V+x\frac{dV}{dx} \;\;\;\ldots (3)\] From (1), (2) and (3) \[V+x\frac{dV}{dx}=\cot V+V\] \[x\frac{dV}{dx}=\cot V\] Separating variables, \[\tan V dV=\frac{1}{x}dx\] Integrating, \[\int\tan V dV=\int\frac{1}{x}dx+\ln C\] $\ln C$ is constant of integration \[-\ln |\cos V|=\ln |x|+\ln |C|\] \[\frac{1}{\cos V}=Cx\] \[V=\cos^{-1}\left(\frac{1}{Cx}\right)\] From (2) \[\frac{y}{x}= \cos^{-1}\left(\frac{1}{Cx}\right) \] \[y=x \cos^{-1}\left(\frac{1}{Cx}\right) \] Hence general solution of (1) is \[y(x)=x \cos^{-1}\left(\frac{1}{Cx}\right) \]
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