If $y=y(x)$ is the solution curve of the differential equation $x^{2} d y+\left(y-\frac{1}{x}\right) d x=0 ; x>0$, and $y(1)=1$, then $\mathrm{y}\left(\frac{1}{2}\right)$ is equal to:

(A) $3+\frac{1}{\sqrt{\mathrm{e}}}$

(B) $\frac{3}{2}-\frac{1}{\sqrt{\mathrm{e}}}$

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Differential Equations

y = y (x)

is the solution curve of the differential equation

x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0

, and

y (1) = 1

, then

y (\frac{1}{2})

is equal to:

Answer: (D)

Writing the equation as

\frac{d y}{d x} + \frac{y}{x ^{2}} = \frac{1}{x ^{3}}

, the integrating factor is

e^{\int x^{- 2} d x} = e^{- 1/ x}

, so

\frac{d}{d x} (y e^{- 1/ x}) = \frac{e ^{- 1/ x}}{x ^{3}} .

With

t = \frac{1}{x}

\int \frac{e ^{- 1/ x}}{x ^{3}} d x = - \int t e^{- t} d t = (t + 1) e^{- t} = (1 + \frac{1}{x}) e^{- 1/ x}

, giving

y e^{- 1/ x} = (1 + \frac{1}{x}) e^{- 1/ x} + C

. From

y (1) = 1

e^{- 1} = 2 e^{- 1} + C

, so

C = - e^{- 1}

. At

x = \frac{1}{2}

y e^{- 2} = 3 e^{- 2} - e^{- 1}

, hence

y = 3 - e

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option