The general solution of the differential equa | vedclass

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(A) Given differential equation: $e^{y-x} \frac{dy}{dx} = y \left( \frac{\sin x + \cos x}{1 + y \log y} \right)$ Rearranging the terms: $\frac{e^y}{e^

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$e^y \log y = e^x \sin x + c$,where $c$ is a constant of integration.

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(A) Given differential equation: $e^{y-x} \frac{dy}{dx} = y \left( \frac{\sin x + \cos x}{1 + y \log y} \right)$ Rearranging the terms: $\frac{e^y}{e^x} \frac{dy}{dx} = \frac{y}{1 + y \log y} (\sin x + \cos x)$ Separating the variables: $\frac{e^y (1 + y \log y)}{y} dy = e^x (\sin x + \cos x) dx$ Simplifying the left side: $e^y \left( \log y + \frac{1}{y} \right) dy = e^x (\sin x + \cos x) dx$ Integrating both sides: $\int e^y \left( \log y + \frac{1}{y} \right) dy = \int e^x (\sin x + \cos x) dx$ Using the identity $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,we get: $e^y \log y = e^x \sin x + c$.

The general solution of the differential equation $e^{y-x} \frac{dy}{dx} = y \left( \frac{\sin x + \cos x}{1 + y \log y} \right)$ is

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