MCQ
The general solution of the differential equation ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$ is
- A$({e^y} + 1)\cos x = K$
- B$({e^y} + 1){\rm{cosec}}\,x = K$
- ✓$({e^y} + 1)\sin x = K$
- DNone of these
==> $\log ({e^y} + 1) + \log \sin x = \log K$ ==> $({e^y} + 1)\sin x = K$.
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