A) \[-{{(n-1)}^{2}}y\]
B) \[{{(n-1)}^{2}}y\]
C) \[-{{n}^{2}}y\]
D) \[{{n}^{2}}y\]
Correct Answer: A
Solution :
We have\[,\]\[y={{x}^{n-1}}\log x\] ... (i) On differentiating Eq. (i), w.r.t.\[x,\] we get \[{{y}_{1}}={{x}^{n-1}}\cdot \frac{1}{x}+(n-1){{x}^{n-2}}\log x\] \[\Rightarrow \] \[x{{y}_{1}}={{x}^{n-1}}+(n-1){{x}^{n-1}}\log x\] \[\Rightarrow \] \[x{{y}_{1}}={{x}^{n-1}}+(n-1)y\][using Eq. (i)] ...(ii) Again differentiating with respect to x, we get \[x{{y}_{2}}+{{y}_{1}}=(n-1){{x}^{n-2}}+(n-1){{y}_{1}}\] \[\Rightarrow \] \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}=(n-1){{x}^{n-1}}+(n-1)x{{y}_{1}}\] \[\Rightarrow \] \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}=(n-1)[x{{y}_{1}}-(n-1)y]\] \[+(n-1)x{{y}_{1}}\][using Eq. (ii)] \[\Rightarrow \]\[{{x}^{2}}{{y}_{2}}+[1-n+1-n+1]x{{y}_{1}}=-{{(n-1)}^{2}}y\] \[\Rightarrow \]\[{{x}^{2}}{{y}_{2}}+(3-2n)x{{y}_{1}}=-{{(n-1)}^{2}}y\]
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