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

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$\begin{aligned} & \sin ^{-1}\left(\frac{d y}{d x}ight)=x+y \ & 	herefore \frac{d y}{d x}=\sin (x+y) \ & 	ext { Put } x+y=t \quad \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \ & 	herefore \frac{d t}{d x}-1=\sin t \Rightarrow \frac{d t}{d x}=1+\sin t \ & 	herefore \int \frac{d t}{1+\sin t}=\int d x \ & 	herefore \int \frac{(1-\sin t) d t}{1-\sin } t=\int d x \Rightarrow \int \frac{1-\sin t}{\cos ^2 t} d t=\int d x \ & \Rightarrow \int\left(\sec ^2 t-\operatorname{sect} 	an tight) d t=\int d x \ & 	herefore 	an t-\sec t=x+c \ & 	herefore 	an (x+y)-\sec (x+y)=x+c\end{aligned}$

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

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