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प्रश्न
If \[y = \sqrt{x + \sqrt{x + \sqrt{x + . . . to \infty ,}}}\] prove that \[\frac{dy}{dx} = \frac{1}{2 y - 1}\] ?
उत्तर
\[\text{ We have, y } = \sqrt{x + \sqrt{x + \sqrt{x + . . . to \infty}}}\]
\[ \Rightarrow y = \sqrt{x + y}\]
\[\text{ Squaring both sides, we get,} \]
\[ y^2 = x + y\]
\[ \Rightarrow 2y \frac{dy}{dx} = 1 + \frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left( 2y - 1 \right) = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2y - 1}\]
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Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
