MCQ
The solution of the equation ${\sin ^{ - 1}}\left( {\frac{{dy}}{{dx}}} \right) = x + y$ is
- A$\tan (x + y) + \sec (x + y) = x + c$
- ✓$\tan (x + y) - \sec (x + y) = x + c$
- C$\tan (x + y) + \sec (x + y) + x + c = 0$
- DNone of these
✓
Answer
Correct option: B.
$\tan (x + y) - \sec (x + y) = x + c$
b
(b) Here $\frac{{dy}}{{dx}} = \sin (x + y)$
(b) Here $\frac{{dy}}{{dx}} = \sin (x + y)$
Now put $x + y = v$ and $\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$
Therefore $\frac{{dy}}{{dx}} = \sin (x + y)$ reduces to $\frac{{dv}}{{1 + \sin v}} = dx$
Now on integrating both the sides, we get
$\tan v - \sec v = x + c$ or $\tan (x + y) - \sec (x + y) = x + c$.
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