MCQ
$(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ નો ઉકેલ મેળવો.
- 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$\log \left( {\frac{x}{y}} \right) = xy + k$
$\frac{{ydx + xdy}}{{{x^2}{y^2}}} + \frac{{dx}}{x} - \frac{{dy}}{y} = 0$. On integrating, we get
$ - \frac{1}{{xy}} + \log x - \log y = k$ ==> $\log \frac{x}{y} = \frac{1}{{xy}} + k$.
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