MCQ
સમીકરણ $({e^x} + 1)ydy = (y + 1){e^x}dx$ નો ઉકેલ મેળવો.
- A$c(y + 1)({e^x} + 1) + {e^y} = 0$
- B$c(y + 1)({e^x} - 1) + {e^y} = 0$
- C$c(y + 1)({e^x} - 1) - {e^y} = 0$
- ✓$c(y + 1)({e^x} + 1) = {e^y}$
==> $\left( {\frac{y}{{y + 1}}} \right)dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$ ==> $\left[ {1 - \frac{1}{{y + 1}}} \right]dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$
==> $\int_{}^{} {\left\{ {1 - \frac{1}{{y + 1}}} \right\}} dy = \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}} dx$
==> $y = \log (y + 1) + \log ({e^x} + 1) + \log c$ or ${e^y} = c(y + 1)({e^x} + 1)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.