MCQ
The solution of differential equation $x\frac{{dy}}{{dx}} + y = {y^2}$ is
- ✓$y = 1 + cxy$
- B$y = \log \{ cxy\} $
- C$y + 1 = cxy$
- D$y = c + xy$
==> $\frac{{dy}}{{{y^2} - y}} = \frac{{dx}}{x}$ ==> $\left[ {\frac{1}{{y - 1}} - \frac{1}{y}} \right]dy = \frac{{dx}}{x}$
On integrating, we get $\log (y - 1) - \log y = \log x + \log c$
==> $\frac{{y - 1}}{y} = xc$ ==> $y = 1 + cxy$.
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