Question

The general solution of the differential equation ${\sin }^{-1}\left(\frac{dy}{dx}\right)=x+y$ is (where $c$ is a constant of integration)

• A. $\sin (x-y)=(x+c)\cos (x+y)+1$
• B. $\sin (x-y)=(x+c)\cos (x-y)+1$
• C. $\sin (x+y)=(x+c)\cos (x+y)+1$
• D. $\sin (x+y)=(x+c)\cos (x-y)+1$

Answer 1

The correct option is C $\sin (x+y)=(x+c)\cos (x+y)+1$

$\frac{dy}{dx}=\sin (x+y)$
Putting $x+y=t$
$\frac{dy}{dx}=\frac{dt}{dx}-1\Rightarrow \frac{dt}{dx}=1+\sin t\Rightarrow \frac{dt}{1+\sin t}=dx$

Integrating both sides,we get

$\int \frac{dt}{1+\sin t}=\int dx$
$\Rightarrow \int \frac{1-\sin t}{{\cos }^{2}t}dt=x+c$
$\Rightarrow \int ({\sec }^{2}t-\sec t\tan t)dt=x+c$
$\tan t-\sec t=x+c$
$\Rightarrow -\frac{1-\sin t}{\cos t}=x+c$
$\Rightarrow \sin t-1=x\cos t+c\cos t$
Substituting the value of $t$
$\sin (x+y)=x\cos (x+y)+c\cos (x+y)+1$