Question

The solution of the equation $\sin ^ { - 1 } \left( \frac { d y } { d x } \right) = x + y$ is

• A. $\tan ( x + y ) + \sec ( x + y ) = x + c$
• B. $\tan ( x + y ) - \sec ( x + y ) = x + c$
• C. $\tan ( x + y ) + \sec ( x + y ) + x + c = 0$
• D. None of these

Answer 1

Answer: (B)

$\tan ( x + y ) - \sec ( x + y ) = x + c$

Answer 2

Here $\frac { d y } { d x } = \sin ( x + y )$

Now put $x + y = v$ and $\frac { d y } { d x } = \frac { d v } { d x } - 1$

Therefore $\frac { d y } { d x } = \sin ( x + y )$reduces to $\frac { d v } { 1 + \sin v } = d x$

Now on integrating both the sides, we get

$\tan v - \sec v = x + c$ or $\tan ( x + y ) - \sec ( x + y ) = x + c$.