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Question
If \[y \sqrt{x^2 + 1} = \log \left( \sqrt{x^2 + 1} - x \right)\] ,Show that \[\left( x^2 + 1 \right) \frac{dy}{dx} + xy + 1 = 0\] ?
Solution
\[\text{ We have, y} \sqrt{x^2 + 1} = \log\left( \sqrt{x^2 + 1} - x \right)\]
Differentiating with respect to x, we get,
\[\Rightarrow \frac{d}{dx}\left( y\sqrt{x^2 + 1} \right) = \frac{d}{dx}\log\left( \sqrt{x^2 + 1} - x \right) \left[ \text{ using product rule and chain rule } \right]\]
\[ \Rightarrow y\frac{d}{dx}\left( \sqrt{x^2 + 1} \right) + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)} \times \frac{d}{dx}\left( \sqrt{x^2 + 1} - x \right)\]
\[ \Rightarrow \frac{y}{2\sqrt{x^2 + 1}} \times \frac{d}{dx}\left( x^2 + 1 \right) + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)} \times \left[ \frac{1}{2\sqrt{x^2 + 1}}\frac{d}{dx}\left( x^2 + 1 \right) - 1 \right]\]
\[ \Rightarrow \frac{2xy}{2\sqrt{x^2 + 1}} + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)}\left[ \frac{2x}{2\sqrt{x^2 + 1}} - 1 \right]\]
\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \left[ \frac{1}{\sqrt{x^2 + 1} - x} \right]\left[ \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right] - \frac{xy}{\sqrt{x^2 + 1}}\]
\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{- 1}{\sqrt{x^2 + 1}} - \frac{xy}{\sqrt{x^2 + 1}}\]
\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{- \left( 1 + xy \right)}{\sqrt{x^2 + 1}}\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{dy}{dx} = - \left( 1 + xy \right)\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{dy}{dx} + 1 + xy = 0\]
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