A) \[y=x\log x+x+2\]
B) \[y=x\log x-x+2\]
C) \[y=x\log x+x\]
D) \[y=x\log x-x\]
Correct Answer: B
Solution :
\[x\frac{{{d}^{2}}y}{d{{x}^{2}}}=1\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{x}\]Þ \[\frac{dy}{dx}=\log x+{{c}_{1}}\] Þ \[y=x\log x-x+{{c}_{1}}x+{{c}_{2}}\] (on integrating twice) Given \[y=1\]and \[\frac{dy}{dx}=0\]at \[x=1\]Þ \[{{c}_{1}}=0\]and \[{{c}_{2}}=2\] Therefore, the required solution is \[y=x\log x-x+2\].
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