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The solution of the differential equation $\sin ^{-1}\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=x+\mathrm{y}$ is
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Verified Answer
The correct answer is:
$x=\tan (x+y)-\sec (x+y)+c$
We have $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$
$$
\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})
$$
Put $\quad x+y=t \Rightarrow y=t-x \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1$
$$
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}=1+\sin \mathrm{t} \Rightarrow \int \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\int \mathrm{dx}
$$
$\int \frac{(1-\sin t)}{(1+\sin t)(1-\sin t)} d t=\int d x \Rightarrow x=\int \frac{1-\sin t}{1-\sin ^{2} t} d t$
$x=\int \frac{1-\sin t}{\cos ^{2} t} d t=\int\left(\sec ^{2} t-\sec t \tan t\right) d t$
$x=\tan t-\sec t+c$
$x=\tan (x+y)-\sec (x+y)+c$
$$
\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})
$$
Put $\quad x+y=t \Rightarrow y=t-x \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1$
$$
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}=1+\sin \mathrm{t} \Rightarrow \int \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\int \mathrm{dx}
$$
$\int \frac{(1-\sin t)}{(1+\sin t)(1-\sin t)} d t=\int d x \Rightarrow x=\int \frac{1-\sin t}{1-\sin ^{2} t} d t$
$x=\int \frac{1-\sin t}{\cos ^{2} t} d t=\int\left(\sec ^{2} t-\sec t \tan t\right) d t$
$x=\tan t-\sec t+c$
$x=\tan (x+y)-\sec (x+y)+c$
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