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. (a) `-e/2`
B. (b) `-e^(2)/2`
C. (c) `e/4`
D. (d) `e^(2)/4`

Answer

(c)

Explanation

Given differential equation is
`xdx/dx+y=x log_(e) x, (x gt 1)`
` rArr dy/dx +1/xy=log_(e)x`
Which is a linear differential equation.
So, `if=e^(int 1/xdx)=e^(log_(e)x)=x`
Now, solution of differential Eq. (i), is
`yxxx=int (log_(e)x)x dx +C`
`rArr yx=x^(2)/2log_(e)x- intx^(2)/2xx1/xdx+C`
[using integration by parts]
`rArr yx=x^(2)/2log_(e)x- x^(2)/4+C ... (ii)`
Given that, `2y(2) = log_(e)4-1`
On substituting, x=2, in Eq. (ii),
we get
`2y(2)=4/2log_(e)2-4/4+C,`
[where, y (2) represents value of y at x =2]
`rArr 2y(2)=log_(e)4-1+C ...(iv)`
`[therefore m log a=log a^(m)]`
From Eqs. (iii) and (iv), we get
`C = 0`
So, required solution is
`yx=x^(2)/2log_(e)x-x^(2)/4`
Now, at `x=e, ey(e) =e^(2)/2log_(e) e-e^(2)/4`
[where, y(e) represents value of y at x =e]
`rArr doty(e)=e/4 [therefore log_(e) e=1].`