A) \[{{n}^{2}}y\]
B) \[-{{n}^{2}}y\]
C) \[-y\]
D) \[2{{x}^{2}}y\]
Correct Answer: A
Solution :
We have\[y={{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] Let \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{y}_{2}}\] and \[\frac{dy}{dx}={{y}_{1}}\] on differentiating Eq. (i), we get \[\frac{dy}{dx}=n{{[x+\sqrt{1+{{x}^{2}}}]}^{n-1}}\]\[\left( 1+\frac{x}{\sqrt{{{x}^{2}}+1}} \right)\] \[=\frac{n{{[x+\sqrt{1+{{x}^{2}}}]}^{n}}}{\sqrt{1+{{x}^{2}}}}\] Or \[\frac{dy}{dx}=\frac{ny}{\sqrt{1+{{x}^{2}}}}\] \[y_{1}^{2}{{(1+x)}^{2}}={{n}^{2}}{{y}^{2}}\] Again differentiating, we get \[2{{y}_{1}}{{y}_{2}}(1+{{x}^{2}})={{n}^{2}}{{y}^{2}}\] Dividing by \[2{{y}_{1}}\] \[{{y}_{1}}(1+{{x}^{2}})+x{{y}_{1}}={{n}^{2}}y\] Or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}{{(1+x)}^{2}}+x\frac{dy}{dx}={{n}^{2}}y\]
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