A) \[\frac{{{e}^{2}}}{4}\]
B) \[-\frac{{{e}^{2}}}{2}\]
C) \[-\frac{e}{2}\]
D) \[\frac{e}{4}\]
Correct Answer: D
Solution :
Here, \[\frac{dy}{dx}+\frac{y}{x}={{\log }_{e}}x\] \[\therefore \]\[I.F.={{e}^{\int_{{}}^{{}}{\frac{1}{x}}dx}}=x\] The required solution is \[xy=\int_{{}}^{{}}{{{\log }_{e}}}x.xdx+C\] \[\Rightarrow \]\[xy=\frac{{{x}^{2}}}{2}{{\log }_{e}}x-\frac{{{x}^{2}}}{4}+C\] ?(i) \[\because \]\[2y(2)=lo{{g}_{e}}4-1\Rightarrow 2y(2)=2lo{{g}_{e}}2-1\] From (i), C = 0 So,\[y=\frac{x}{2}{{\log }_{e}}x-\frac{x}{4}\Rightarrow y(e)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}\]
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