The general solution of the differential equa | vedclass

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Question

(D) Given differential equation: $\left(\frac{y}{x}\right) \cos \left(\frac{y}{x}\right) dx = \left[\frac{x}{y} \sin \left(\frac{y}{x}\right) + \cos \

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$y \sin \left(\frac{y}{x}\right) = k$

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(D) Given differential equation: $\left(\frac{y}{x}\right) \cos \left(\frac{y}{x}\right) dx = \left[\frac{x}{y} \sin \left(\frac{y}{x}\right) + \cos \left(\frac{y}{x}\right)\right] dy$. Rearranging,we get $\frac{dy}{dx} = \frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)}{\frac{x}{y} \sin \left(\frac{y}{x}\right) + \cos \left(\frac{y}{x}\right)}$. Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{v \cos v}{\frac{1}{v} \sin v + \cos v} = \frac{v^2 \cos v}{\sin v + v \cos v}$. Then $x \frac{dv}{dx} = \frac{v^2 \cos v}{\sin v + v \cos v} - v = \frac{v^2 \cos v - v \sin v - v^2 \cos v}{\sin v + v \cos v} = \frac{-v \sin v}{\sin v + v \cos v}$. Separating variables: $\frac{\sin v + v \cos v}{v \sin v} dv = -\frac{dx}{x}$. Integrating both sides: $\int \left( \frac{1}{v} + \cot v \right) dv = -\int \frac{dx}{x}$. $\ln |v| + \ln |\sin v| = -\ln |x| + \ln |k|$. $\ln |v \sin v x| = \ln |k| \Rightarrow v \sin v x = k$. Since $v = \frac{y}{x}$,we have $\frac{y}{x} \sin \left(\frac{y}{x}\right) x = k$,which simplifies to $y \sin \left(\frac{y}{x}\right) = k$.

The general solution of the differential equation $\left(\frac{y}{x}\right) \cos \left(\frac{y}{x}\right) dx - \left[\left(\frac{x}{y}\right) \sin \left(\frac{y}{x}\right) + \cos \left(\frac{y}{x}\right)\right] dy = 0$ is:

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