The general solution of the equation ({e^y} + | vedclass

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(C) Given differential equation is $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$. Rearranging the terms to separate the variables $x$ and $y$,we g

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$({e^y} + 1)\sin x = c$

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(C) Given differential equation is $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$. Rearranging the terms to separate the variables $x$ and $y$,we get: $\frac{{e^y}}{{e^y} + 1} \, dy + \frac{{\cos x}}{{\sin x}} \, dx = 0$. Integrating both sides: $\int \frac{{e^y}}{{e^y} + 1} \, dy + \int \frac{{\cos x}}{{\sin x}} \, dx = C$. Let $u = {e^y} + 1$,then $du = {e^y} \, dy$. The integral becomes $\int \frac{1}{u} \, du + \int \cot x \, dx = C$. This results in $\ln({e^y} + 1) + \ln(\sin x) = \ln c$. Using the property $\ln a + \ln b = \ln(ab)$,we get $\ln(({e^y} + 1)\sin x) = \ln c$. Taking the exponential of both sides,we obtain $({e^y} + 1)\sin x = c$.

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

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