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AP EAPCET 19-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 65 of 160

Marks: +1, -0

If

\log (\sqrt{1+x^{2}}-x)=y(\sqrt{1+x^{2}})

(1+x^{2}) \frac{dy}{dx}+xy=

0
1
2
-1

Solution:

Given,

y \sqrt{1+x^{2}}=\log [\sqrt{1+x^{2}}-x]

On differentiating both sides with respect to x, we get,

y \cdot \frac{1}{2 \sqrt{1 \div x^{2}}} \cdot 2x+\sqrt{1+x^{2}} \frac{dy}{dx}

=\frac{1}{\sqrt{1+x^{2}-x}}\left\{\frac{1}{2 \sqrt{1+x^{2}}} \cdot 2x-1\right\}

\Rightarrow \frac{xy}{\sqrt{1+x^{2}}}+(\sqrt{1+x^{2}}) \frac{dy}{dx}

=\frac{1}{\sqrt{1+x^{2}}-x} \cdot\left\{\frac{x}{\sqrt{1+x^{2}}}-1\right\}

\Rightarrow \frac{xy}{\sqrt{1+x^{2}}}+\sqrt{1+x^{2}} \frac{dy}{dx}

=\frac{-(\sqrt{1+x^{2}}-x)}{\sqrt{1+x^{2}}-x} \cdot\frac{1}{\sqrt{1+x^{2}}}

xy+(1+x^{2}) \frac{dy}{dx}=-1

or

(1+x^{2}) \frac{dy}{dx}+xy=-1

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