(1+2²*)dx =0dy + 2Find the gereral solution.

To solve the differential equation: 1+2eyxdy+2eyx1−yxdx=01+2eyxdy=−2eyx1−yxdx⇒dydx=2eyx−1+yx1+2eyx

Answered: 는 (1-5)~ dx =C 1+2e dy + 2 Find the…

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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Question

(1+2²*)
dx =0
dy + 2
Find the gereral solution.

Expert Solution

Step 1

To solve the differential equation:

$\begin{array}{l} (1 + 2 e^{\frac{y}{x}}) d y + 2 e^{\frac{y}{x}} (1 - \frac{y}{x}) d x = 0 \\ (1 + 2 e^{\frac{y}{x}}) d y = - 2 e^{\frac{y}{x}} (1 - \frac{y}{x}) d x \\ \Rightarrow \frac{d y}{d x} = \frac{2 e^{\frac{y}{x}} (- 1 + \frac{y}{x})}{1 + 2 e^{\frac{y}{x}}} \end{array}$

Step 2

Consider:

$\begin{array}{l} \frac{y}{x} = v \Rightarrow y = v x \\ \Rightarrow \frac{d y}{d x} = \frac{d (v x)}{d x} = v + x \frac{d v}{d x} \\ Thus, \\ \frac{d y}{d x} = \frac{2 e^{\frac{y}{x}} (- 1 + \frac{y}{x})}{1 + 2 e^{\frac{y}{x}}} is \\ v + x \frac{d v}{d x} = \frac{2 e^{v} (v - 1)}{1 + 2 e^{v}} \Rightarrow x \frac{d v}{d x} = \frac{2 e^{v} (v - 1)}{1 + 2 e^{v}} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{2 e^{v} v - 2 e^{v} - v - 2 v e^{v}}{1 + 2 e^{v}} = \frac{- (2 e^{v} + v)}{1 + 2 e^{v}} \\ \Rightarrow \frac{1 + 2 e^{v}}{2 e^{v} + v} d v = \frac{d x}{x} \\ \Rightarrow \int \frac{1 + 2 e^{v}}{2 e^{v} + v} d v = \int \frac{d x}{x} \\ \ln (2 e^{v} + v) = \ln x + C; where C is an arbitrary constant \\ \Rightarrow 2 e^{v} + v = e^{\ln x + C} = e^{\ln x} e^{C} = x e^{C} \\ \Rightarrow 2 e^{v} + v = A x; where A = e^{C} is an arbitrary constant \end{array}$

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