A) \[{{n}^{2}}y\]
B) \[-{{n}^{2}}y\]
C) \[-y\]
D) \[2{{x}^{2}}y\]
Correct Answer: A
Solution :
[a] \[y={{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] \[\frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n-1}}\left( 1+\frac{1}{2}{{(1+{{x}^{2}})}^{-1/2}}.2x \right);\] \[\frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n-1}}\frac{(\sqrt{1+{{x}^{2}}}+x)}{\sqrt{1+{{x}^{2}}}}=\frac{n{{(\sqrt{1+{{x}^{2}}}+x)}^{n}}}{\sqrt{1+{{x}^{2}}}}\] or \[\sqrt{1+{{x}^{2}}}\frac{dy}{dx}=ny\] or \[\sqrt{1+{{x}^{2}}}{{y}_{1}}=ny({{y}_{1}}=\frac{dy}{dx})\] Squaring, \[(1+{{x}^{2}}){{y}_{1}}^{2}.\,{{n}^{2}}{{y}^{2}}\] Differentiating, \[(1+{{x}^{2}})2{{y}_{1}}{{y}_{2}}+{{y}_{1}}^{2}.2x={{n}^{2}}.2y{{y}_{1}}\] \[(Here,{{y}_{2}}=\frac{{{d}^{2}}y}{d{{x}^{2}}})\] or \[(1+{{x}^{2}}){{y}_{2}}+x{{y}_{1}}={{n}^{2}}y\]
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