Let y=y(x) be the solution of the differential equation, x(dxdy)+y=xlogex, (x>1) if 2y(2)=loge4−1, then y(e) is equal to: (a) −(2e) (b) −(2e2) (c) 4e (d) 4e2

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Let {y}={y}{({x})} be the solution of the differential equatio

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Let $y = y (x)$ be the solution of the differential equation, $x (\frac{d y}{d x}) + y = x lo g_{e} x$ , $(x > 1)$ if $2 y (2) = lo g_{e} 4 - 1$ , then $y (e)$ is equal to: (a) $- (\frac{e}{2})$ (b) $- (\frac{e ^{2}}{2})$ (c) $\frac{e}{4}$ (d) $\frac{e ^{2}}{4}$

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Let

y = y (x)

be the solution of the differential equation,

x (\frac{d y}{d x}) + y = x lo g_{e} x

(x > 1)

2 y (2) = lo g_{e} 4 - 1

, then

y (e)

is equal to: (a)

- (\frac{e}{2})

(b)

- (\frac{e ^{2}}{2})

(c)

\frac{e}{4}

(d)

\frac{e ^{2}}{4}

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Question Text	Let $y = y (x)$ be the solution of the differential equation, $x (\frac{d y}{d x}) + y = x lo g_{e} x$ , $(x > 1)$ if $2 y (2) = lo g_{e} 4 - 1$ , then $y (e)$ is equal to: (a) $- (\frac{e}{2})$ (b) $- (\frac{e ^{2}}{2})$ (c) $\frac{e}{4}$ (d) $\frac{e ^{2}}{4}$
Answer Type	Video solution: 1
Upvotes	150