The equation of the curve passing through lef | vedclass

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(A) Given differential equation is $(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0$. Rearranging the terms,we get $\frac{e^y}{e^y + 1} \, dy = -\frac{\

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$y = \log_e(\operatorname{cosec} x - 1)$

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(A) Given differential equation is $(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0$. Rearranging the terms,we get $\frac{e^y}{e^y + 1} \, dy = -\frac{\cos x}{\sin x} \, dx$. 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 $\int \frac{1}{u} \, du = -\ln|\sin x| + C$. So,$\ln(e^y + 1) = -\ln|\sin x| + C$,which simplifies to $\ln(e^y + 1) + \ln|\sin x| = C$. This gives $\ln((e^y + 1) \sin x) = C$,or $(e^y + 1) \sin x = K$ (where $K = e^C$). The curve passes through $\left(\frac{\pi}{6}, 0\right)$,so substitute $x = \frac{\pi}{6}$ and $y = 0$: $(e^0 + 1) \sin(\frac{\pi}{6}) = K \implies (1 + 1) \cdot \frac{1}{2} = K \implies K = 1$. Thus,$(e^y + 1) \sin x = 1$,which means $e^y + 1 = \frac{1}{\sin x} = \operatorname{cosec} x$. Therefore,$e^y = \operatorname{cosec} x - 1$,and $y = \log_e(\operatorname{cosec} x - 1)$.

The equation of the curve passing through $\left(\frac{\pi}{6}, 0\right)$ and satisfying the differential equation $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ is:

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