Question

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

• A. $\left\{\frac{-2}{(1+\tan \frac{x+y}{2})}\right\}=x+c.$ .
• B. $\left\{\frac{2}{(1+\tan \frac{x+y}{2})}\right\}=x+c.$ .
• C. $\left\{\frac{-2}{(1-\tan \frac{x+y}{2})}\right\}=x+c.$ .
• D. $\left\{\frac{2}{(1-\tan \frac{x+y}{2})}\right\}=x+c.$ .

Answer 1

The correct option is A $\left\{\frac{-2}{(1+\tan \frac{x+y}{2})}\right\}=x+c.$ .

$\frac{dy}{dx}=\sin (x+y)$
put $x+y=v\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$
$\therefore \frac{dv}{dx}=1+\sin v={\cos }^2\frac{v}{2}+{\sin }^2\frac{v}{2}+2\sin \frac{v}{2}\cos \frac{v}{2}$
$\Rightarrow \frac{dv}{{(\cos \frac{v}{2}+\sin \frac{v}{2})}^2}=dx$
or $\frac{{\sec }^2\frac{v}{2}}{{(1+\tan \frac{v}{2})}^2}dv=dx$
Integrating we get,
$\left\{\frac{-2}{(1+\tan \frac{v}{2})}\right\}=x+c.$