The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$Rearranging the terms to separate the variables:${e^y}\frac{{dy}}

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

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(C) Given the differential equation: ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$Rearranging the terms to separate the variables:${e^y}\frac{{dy}}{{dx}} = -({e^y} + 1)\cot x$$\frac{{{e^y}}}{{{e^y} + 1}}dy = -\cot x dx$Integrating both sides:$\int \frac{{{e^y}}}{{{e^y} + 1}}dy = -\int \cot x dx$Let $u = {e^y} + 1$,then $du = {e^y} dy$. The integral becomes:$\int \frac{1}{u} du = -\int \cot x dx$$\ln|{e^y} + 1| = -\ln|\sin x| + C$$\ln|{e^y} + 1| + \ln|\sin x| = C$Using the property $\ln A + \ln B = \ln(AB)$:$\ln|({e^y} + 1)\sin x| = C$Taking the exponential of both sides,we get:$({e^y} + 1)\sin x = K$ (where $K = e^C$ is a constant).

The general solution of the differential equation ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$ is

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