The solution of (1 + xy)y ,dx + (1 - xy)x ,dy = 0 is - Vidyadip

MCQ

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is

A
$\frac{x}{y} + \frac{1}{{xy}} = k$
✓
$\log \left( {\frac{x}{y}} \right) = \frac{1}{{xy}} + k$
C
$\frac{x}{y} + \frac{1}{{xy}} = k$
D
$\log \left( {\frac{x}{y}} \right) = xy + k$

✓

Answer

Correct option: B.

$\log \left( {\frac{x}{y}} \right) = \frac{1}{{xy}} + k$

b
(b) $ydx + xdy + x{y^2}dx - {x^2}ydy = 0$

$\frac{{ydx + xdy}}{{{x^2}{y^2}}} + \frac{{dx}}{x} - \frac{{dy}}{y} = 0$. On integrating, we get

$ - \frac{1}{{xy}} + \log x - \log y = k$ ==> $\log \frac{x}{y} = \frac{1}{{xy}} + k$.

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