The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$. Rearranging,we get: $\frac{dy}{dx} = \frac{y \sin (y/x)

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$\cos (\frac{y}{x}) = \log |x| + c$

Vedclass · Answer

(A) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$. Rearranging,we get: $\frac{dy}{dx} = \frac{y \sin (y/x) - x}{x \sin (y/x)}$. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting into the equation: $v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v}$. Subtracting $v$ from both sides: $x \frac{dv}{dx} = -\frac{1}{\sin v}$. Separating variables: $\sin v \, dv = -\frac{1}{x} \, dx$. Integrating both sides: $\int \sin v \, dv = -\int \frac{1}{x} \, dx$. $-\cos v = -\log |x| + C_1$,which simplifies to $\cos v = \log |x| + c$. Substituting $v = y/x$ back,we get: $\cos (\frac{y}{x}) = \log |x| + c$.

The general solution of the differential equation $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$ is

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