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The general solution of the differential equation $\frac{d y}{d x}+\frac{1}{\sqrt{1-x^{2}}}=0$ is
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The correct answer is:
$y+\sin ^{-1} x=c$
$\frac{d y}{d x}+\frac{1}{\sqrt{1-x^{2}}}=0$
$\therefore \frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \Rightarrow d y=\frac{-d x}{\sqrt{1-x^{2}}} \Rightarrow \int d y=-\int \frac{d x}{\sqrt{1-x^{2}}}$
$\therefore y=-\sin ^{-1} x+c$
$\therefore \frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \Rightarrow d y=\frac{-d x}{\sqrt{1-x^{2}}} \Rightarrow \int d y=-\int \frac{d x}{\sqrt{1-x^{2}}}$
$\therefore y=-\sin ^{-1} x+c$
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