MCQ
સમીકરણ $({e^y} + 1)\cos xdx + {e^y}\sin xdy = 0$ નો ઉકેલ મેળવો.
- A$({e^y} + 1)\cos x = c$
- B$({e^y} - 1)\sin x = c$
- ✓$({e^y} + 1)\sin x = c$
- Dએકપણ નહી.
==> $\frac{{{e^y}dy}}{{{e^y} + 1}} + \frac{{\cos x}}{{\sin x}}dx = 0$
On integrating both the functions, we get
$\log ({e^y} + 1) + \log (\sin x) = \log c$ ==>$({e^y} + 1)\sin x = c$.
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