The general solution of sin ^{-1}left(frac{d | vedclass

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(D) Given equation: $\sin ^{-1}\left(\frac{d y}{d x}ight)=x+y$ $	herefore \frac{d y}{d x}=\sin (x+y)$ Let $x+y=t$. Differentiating with respect to

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

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(D) Given equation: $\sin ^{-1}\left(\frac{d y}{d x}ight)=x+y$ $	herefore \frac{d y}{d x}=\sin (x+y)$ Let $x+y=t$. Differentiating with respect to $x$,we get $1+\frac{d y}{d x}=\frac{d t}{d x}$,so $\frac{d y}{d x}=\frac{d t}{d x}-1$. Substituting this into the differential equation: $\frac{d t}{d x}-1=\sin t$ $\frac{d t}{d x}=1+\sin t$ Separating variables: $\int \frac{d t}{1+\sin t}=\int d x$ Multiply numerator and denominator by $(1-\sin t)$: $\int \frac{1-\sin t}{1-\sin^2 t} d t=\int d x$ $\int \frac{1-\sin t}{\cos^2 t} d t=\int d x$ $\int (\sec^2 t - \sec t 	an t) d t = \int d x$ Integrating both sides: $	an t - \sec t = x + c$ Substituting $t = x+y$ back: $	an (x+y) - \sec (x+y) = x + c$

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

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