JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x

x > 0

and

y (e^{- 1}) = 0

. Then,

y (e)

is equal to

Answer: (C)

Solution

Rewrite as

\frac{d y}{d x} + \frac{y}{x ln x} = \frac{3}{2 x ^{2}}

. IF

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

y ln x = \int \frac{3 ln x}{2 x ^{2}} d x = - \frac{3 ln x}{2 x} - \frac{3}{2 x} + C .

y (e^{- 1}) = 0

gives

C = 0

. At

x = e

y = - \frac{3}{2 e} - \frac{3}{2 e} = - \frac{3}{e} .

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