QJEEM19Ma3421: JEE Main 2019 Mathematics - Class 12 JEEM19Ma3421 Solution: Let y = y(x) be the solution of the differential equation, x d…

Question

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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 2y(2) = log_e 4

-

1, then y(e) is equal to :

Accepted Answer

\frac{d y}{d x} = \frac{y}{x} = ℓ n x

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

x y = \int x ℓ n x + C

ℓ n x \frac{x^{2}}{2} - \int \frac{1}{x} . \frac{x^{2}}{2}

x y = \frac{x}{2} ℓ n x - \frac{x^{2}}{4} + C,

for

2 y (2) = 2 ℓ n 2 - 1

\Rightarrow

C = 0

y = \frac{x}{2} ℓ n x - \frac{x}{4}

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

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