The solution of e^{y-x} frac{dy}{dx} = frac{y | vedclass

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Question

(C) Given the differential equation: $e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y}$ Separate the variables: $e^y \frac{1 + y \log y

Vedclass · Accepted Answer

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

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(C) Given the differential equation: $e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y}$ Separate the variables: $e^y \frac{1 + y \log y}{y} dy = e^x (\sin x + \cos x) dx$ Simplify the left side: $e^y (\log y + \frac{1}{y}) dy = e^x (\sin x + \cos x) dx$ Integrate both sides: $\int e^y (\log y + \frac{1}{y}) dy = \int e^x (\sin x + \cos x) dx$ Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(y) = \log y$ and $f(x) = \sin x$: $e^y \log y = e^x \sin x + c$ Thus,the correct option is $C$.

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

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