Let $\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $(\mathrm{x} \log \mathrm{x}) \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=2 \mathrm{x} \log \mathrm{x},(\mathrm{x} \geq 1)$. Then $\mathrm{y}(\mathrm{e})$ is equal to :

JEE Main · 2015 · 04 Apr 2015 · Offline · Differential Equations

Let

y (x)

be the solution of the differential equation

(x lo g x) \frac{dy}{dx} + y = 2 x lo g x, (x \geq 1)

. Then

y (e)

is equal to :

Answer: (B)

Put the equation into linear form by dividing through:

\frac{d y}{d x} + \frac{y}{x l n x} = 2

. The integrating factor is

e^{\int \frac{d x}{x l n x}} = e^{l n (l n x)} = ln x

, which packages the left side as

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

. Integrating,

y ln x = \int 2 ln x d x = 2 (x ln x - x) + c

; the condition at

x = 1

, where

ln x = 0

, forces

0 = 2 (0 - 1) + c

, so

c = 2

. Evaluating at

x = e

y \cdot 1 = 2 (e - e + 1) = 2

, hence

y (e) = 2

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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