Let y = y(x) be the solution of the differential equation `x dy/dx+y=xlog_ex,(xgt1)." If " 2y(2)=log_e4-1," then "y(e)` is equal to

Options

A. `(e )/(4)`
B. `-(e^2)/(2)`
C. `(e^2)/(4)`
D. `-(e )/(2)`

Answer

A

Explanation

`(dy)/(dx)+(y)/(x)=log_(e)x`
I.F.`=_eint(1)/(2)dx =e^(lnx)=x`
Solution is `yx=int x.log_(e) x.dx+C=yx=(x^2)/(2)log_(e)x-int(1)/(x).(x^2)/(2)dx+C`
`yx=(x^2)/(2) log_(e)x-(1)/(2).(x^2)/(2)+C`.
At `x=2`
`2y(2)=2log_(e)2-1+C rArr C=0`
`yx =(x^2)/(2)log_(e)x-(x^2)/(4)`
At `x=e`
ey(e)`=(e^2)/(2)-(e^2)/(4)=(e^2)/(4)`
`y(e)=e//4`.