Let $y=y(x)$ be the solution of the differential equation, $x \frac{d y}{d x}+y=x \log _{e} x,(x>1)$. If $2 y(2) =\log _{\mathrm{e}} 4-1$, then $\mathrm{y}(\mathrm{e})$ is equal to:

(B) $-\frac{\mathrm{e}^{2}}{2}$

(D) $\frac{\mathrm{e}^{2}}{4}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Differential Equations

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:

Answer: (C)

Divide through by

x

to reach the linear form

\frac{d y}{d x} + \frac{y}{x} = ln x

, whose integrating factor is

e^{\int \frac{1}{x} d x} = x

. Multiplying back, the left side is

\frac{d}{d x} (x y)

, so

x y = \int x ln x d x = \frac{x ^{2} l n x}{2} - \frac{x ^{2}}{4} + C

. The condition

2 y (2) = ln 4 - 1

means

y (2) = \frac{l n 4 - 1}{2}

; substituting

x = 2

shows

C = 0

. Hence

y = \frac{x l n x}{2} - \frac{x}{4}

, and at

x = e

y (e) = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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