Solve cos (2) - [sin[sin (?) + cos (:)]* =Ydy 0dzO a.O b.POy cos(²-) =y sin ( ² ) =OC. y cos () =y sin () == c= Cz= cx

Given differential equation is yxcosyx-xysinyx+cosyxdydx=0 (1) Simplifying (1), we get…

Answered: Solvecos (2) O a. O b. 2 с. d. y cos Y…

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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Related questions

Question

Solve cos (2) - [sin
[sin (?) + cos (:)]* =
Y
dy 0
dz
O a.
O b.
PO
y cos
(²-) =
y sin ( ² ) =
OC. y cos () =
y sin () == c
= Cz
= cx

Expert Solution

Step 1

Given differential equation is

$\frac{y}{x} \cos (\frac{y}{x}) - [\frac{x}{y} \sin (\frac{y}{x}) + \cos (\frac{y}{x})] \frac{d y}{d x} = 0$ (1)

Simplifying (1), we get

$\frac{d y}{d x} = \frac{\frac{y}{x} \cos (\frac{y}{x})}{\frac{x}{y} \sin (\frac{y}{x}) + \cos (\frac{y}{x})}$ (2)

The above differential equation is homogeneous differential equation.

Let

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

Putting above values in (2)

$v + x \frac{d v}{d x} = \frac{v \cos v}{\frac{1}{v} \sin v + \cos v} \Rightarrow v + x \frac{d v}{d x} = \frac{v^{2} \cos v}{\sin v + v \cos v} \Rightarrow x \frac{d v}{d x} = \frac{v^{2} \cos v}{\sin v + v \cos v} - v \Rightarrow x \frac{d v}{d x} = \frac{v^{2} \cos v - v \sin v - v^{2} \cos v}{\sin v + v \cos v} \Rightarrow x \frac{d v}{d x} = \frac{- v \sin v}{\sin v + v \cos v}$