A) \[-\,({{n}^{2}}+{{a}^{2}})\,{{y}_{n}}\]
B) \[\,({{n}^{2}}-{{a}^{2}})\,{{y}_{n}}\]
C) \[\,({{n}^{2}}+{{a}^{2}})\,{{y}_{n}}\]
D) \[-\,({{n}^{2}}-{{a}^{2}})\,{{y}_{n}}\]
Correct Answer: C
Solution :
Given, \[y={{e}^{a\,{{\sin }^{-1}}x}}\] On differentiating w.r.t. x, we get \[{{y}_{1}}={{e}^{a{{\sin }^{-1}}x}}a\cdot \frac{1}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \] \[{{y}_{1}}\sqrt{1-{{x}^{2}}}=ay\] \[\Rightarrow \] \[(1-{{x}^{2}})y_{1}^{2}={{a}^{2}}{{y}^{2}}\] Again, differentiating w.r.t. x, we get \[(1-{{x}^{2}})2{{y}_{1}}{{y}_{2}}-2xy_{1}^{2}={{a}^{2}}2y{{y}_{1}}\] \[\Rightarrow \] \[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}-{{a}^{2}}y=0\] Using Leibnitzs rule, \[(1-{{x}^{2}}){{y}_{n+2}}+{}^{n}{{C}_{1}}{{y}_{n+1}}(-2x)+{}^{n}{{C}_{2}}{{y}_{n}}(-2)\] \[-x{{y}_{n+1}}-{}^{n}{{C}_{1}}{{y}_{n}}-{{a}^{2}}{{y}_{n}}=0\] \[\Rightarrow \] \[(1-{{x}^{2}}){{y}_{n+2}}+x{{y}_{n+1}}(-2n-1)\] \[+{{y}_{n}}[-n(n-1)-n-{{a}^{2}}]=0\] \[\Rightarrow \] \[(1-{{x}^{2}}){{y}_{n+2}}-(2n+1)x{{y}_{n+1}}=({{n}^{2}}+{{a}^{2}}){{y}_{n}}\]
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