MCQ
$ydx\, + (1 + {x^2}){\tan ^{ - 1}}xdy = 0,$ નો ઉકેલ મેળવો.
- ✓$y{\tan ^{ - 1}}x = c$
- B$x{\tan ^{ - 1}}y = c$
- C$y + {\tan ^{ - 1}}x = c$
- D$x + {\tan ^{ - 1}}y = c$
==> $\int_{}^{} {\frac{{dx}}{{(1 + {x^2}){{\tan }^{ - 1}}x}}} = - \int_{}^{} {\frac{{dy}}{y}} $
==> $\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = \frac{{{c_1}}}{{\log 2}} = c$
==> $\log (y{\tan ^{ - 1}}x) + \log c = 0$ ==>$y {\tan ^{ - 1}}x = c$.
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