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$ ==> $\log \frac{x}{y} = \frac{1}{{xy}} + k$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\sin \left(2 x^{2}\right) \log _{e}\left(\tan x^{2}\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)\right) d x=0$,$0 < x < \sqrt{\frac{\pi}{2}}$ નો ઉકેલ વક્ર $y=y(x)$ છે. જે બિંદુ $\left(\sqrt{\frac{\pi}{6}}, 1\right)$ માંથી પસાર થાય છે. તો $\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|=$ ..............