Answered: (0) 10) dy y+ J 2. 2. dx メーツ X 70

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Step 1

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Given differential equation is

$x \frac{d y}{d x} = y + \sqrt{x^{2} - y^{2}}, x > 0$

Dividing both sides by $x$ , we get

$\frac{d y}{d x} = \frac{y}{x} + \frac{1}{x} \sqrt{x^{2} - y^{2}} \frac{d y}{d x} = \frac{y}{x} + \sqrt{\frac{x^{2} - y^{2}}{x^{2}}} \frac{d y}{d x} = \frac{y}{x} + \sqrt{1 - {(\frac{y}{x})}^{2}} (1)$

which is homogeneous linear differential equation. To solve such equations we substitute

$y = v x \frac{d y}{d x} = v + x \frac{d v}{d x}$

Step 2

So, from (1), we get

$v + x \frac{d v}{d x} = v + \sqrt{1 - v^{2}}$

Cancelling $v$ from both sides,

$x \frac{d v}{d x} = \sqrt{1 - v^{2}} \frac{d v}{\sqrt{1 - v^{2}}} = \frac{d x}{x}$

Integrating both sides , we get

$\sin^{- 1} v = \log x + C [\frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a})] \sin^{- 1} (\frac{y}{x}) = \log x + C \frac{y}{x} = \sin (\log x + C) y = x \sin (logx + C)$

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