Question

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

Answer 1

The correct option is B $\tan (x+y)-\sec (x+y)=x+C$

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 sides, we get
$\tan v-\sec v=x+C$
or $\tan (x+y)-\sec (x+y)=x+C$.