MCQ
The solution of the equation $\frac{{dy}}{{dx}} = y({e^x} + 1)$ is
- A$y + {e^{({e^x} + x + c)}} = 0$
- ✓$\log y = {e^x} + x + c$
- C$\log y + {e^x} = x + c$
- DNone of these
✓
Answer
Correct option: B.
$\log y = {e^x} + x + c$
b
(b) $\frac{{dy}}{{dx}} = y({e^x} + 1)$ ==> $\frac{{dy}}{y} = ({e^x} + 1)dx$
(b) $\frac{{dy}}{{dx}} = y({e^x} + 1)$ ==> $\frac{{dy}}{y} = ({e^x} + 1)dx$
Integrating both sides, we get $\log y = {e^x} + x + c$.
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