MCQ
The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$ is
- A$1 + xy + c(y + x) = 0$
- B$x + y = c(1 - xy)$
- ✓$y - x = c(1 + xy)$
- D$1 + xy = c(x + y)$
Now on integrating both sides, we get
${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}c$==> ${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + c}}{{1 - cx}}} \right)$
==> $y = \frac{{x + c}}{{1 - cx}}$ ==> $y - x = c(1 + xy)$.
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