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
    $y = c + xy$

Answer

Correct option: A.
$y = 1 + cxy$
a
(a) $x\frac{{dy}}{{dx}} + y = {y^2}$ ==> $x\frac{{dy}}{{dx}} = {y^2} - y$

==> $\frac{{dy}}{{{y^2} - y}} = \frac{{dx}}{x}$ ==> $\left[ {\frac{1}{{y - 1}} - \frac{1}{y}} \right]dy = \frac{{dx}}{x}$

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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