A) \[{{x}^{x}}={{e}^{y{{e}^{y}}}}\]
B) \[{{e}^{y}}={{x}^{{{e}^{y}}}}\]
C) \[{{x}^{x}}=y{{e}^{^{y}}}\]
D) None of these
Correct Answer: A
Solution :
[a] \[(1+logx)\frac{dx}{dy}-x\log x={{e}^{y}}\] putting x log \[x\log x=t\Rightarrow (1+logx)dx=dt\] \[\therefore \frac{dt}{dy}-t={{e}^{y}}\] Now, I.F. \[={{e}^{\int{-1dy}}}={{e}^{-y}}\] \[\Rightarrow t{{e}^{-y}}=\int{{{e}^{-y}}{{e}^{y}}dy+C}\] \[\Rightarrow t=C{{e}^{y}}+y{{e}^{y}}\Rightarrow x\log x=(C+y){{e}^{y}},\] Since, \[y(1)=0,\] Then \[0=(C+0)1\Rightarrow C=0\] \[\therefore y{{e}^{y}}=x\log x\Rightarrow {{x}^{x}}={{e}^{y{{e}^{y}}}}\]
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