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