The general solution of left(x frac{dy}{dx} - | vedclass

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(A) Given equation: $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ Divide both sides by $x^2$: $\left(\frac{x \frac{dy}{dx} - y}{x^2}\r

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$e^x(x - 1) + \cos \frac{y}{x} + c = 0$

Vedclass · Answer

(A) Given equation: $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ Divide both sides by $x^2$: $\left(\frac{x \frac{dy}{dx} - y}{x^2}\right) \sin \frac{y}{x} = x e^x$ Let $t = \frac{y}{x}$. Then $\frac{dt}{dx} = \frac{x \frac{dy}{dx} - y}{x^2}$. Substituting this into the equation: $\frac{dt}{dx} \sin t = x e^x$ Integrating both sides with respect to $x$: $\int \sin t \, dt = \int x e^x \, dx$ Using integration by parts for the right side: $-\cos t = x e^x - \int e^x \, dx$ $-\cos t = x e^x - e^x + c$ $-\cos t = e^x(x - 1) + c$ Substituting $t = \frac{y}{x}$ back: $-\cos \frac{y}{x} = e^x(x - 1) + c$ Rearranging gives: $e^x(x - 1) + \cos \frac{y}{x} + c = 0$

The general solution of $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ is

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