Let $\mathrm{x}=\mathrm{x}(\mathrm{y})$ be the solution of the differential equation $y^{2} dx+\left(x-\frac{1}{y}\right) dy=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Differential Equations

Let

x = x (y)

be the solution of the differential equation

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

. If

x (1) = 1

, then

x (\frac{1}{2})

is :

Answer: (C)

Rewrite the equation with

x

as the dependent variable:

y^{2} d x + (x - \frac{1}{y}) d y = 0 ⟹ \frac{d x}{d y} + \frac{1}{y ^{2}} x = \frac{1}{y ^{3}} .

This is linear in

x

, with integrating factor

I.F. = e^{\int \frac{1}{y ^{2}} d y} = e^{- 1/ y} .

x e^{- 1/ y} = \int e^{- 1/ y} \cdot \frac{1}{y ^{3}} d y .

Substitute

t = - \frac{1}{y}

, so

d t = \frac{1}{y ^{2}} d y

and

\frac{1}{y} = - t

; then

\frac{1}{y ^{3}} d y = \frac{1}{y} \cdot \frac{d y}{y ^{2}} = - t d t

. The integral becomes

\int e^{t} (- t) d t = - (t e^{t} - e^{t}) + C = - t e^{t} + e^{t} + C .

Back-substituting

t = - 1/ y

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

Using

x (1) = 1

(i.e.

y = 1, x = 1

e^{- 1} = e^{- 1} + e^{- 1} + C ⟹ C = - e^{- 1} .

Now put

y = \frac{1}{2}

, so

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

x e^{- 2} = 2 e^{- 2} + e^{- 2} - e^{- 1} = 3 e^{- 2} - e^{- 1} .

Dividing by

e^{- 2}

x = 3 - e .

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option