JEE Main 2025 Jan 22 Shift 1, Mathematics Q12: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Jan 22, Shift 1Mathematics

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 :

A
$\frac{1}{2} + e$
B
$\frac{3}{2} + e$
C
$3 - e$
D
$3 + e$

Solution

Rewrite as a linear ODE in $x$ with $y$ as independent variable:

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

Integrating factor: $μ (y) = e^{\int 1/ y^{2} d y} = e^{- 1/ y}$ .

$\frac{d}{d y} [x e^{- 1/ y}] = \frac{1}{y ^{3}} e^{- 1/ y}$ .

Substitute $t = - 1/ y$ , $d t = 1/ y^{2} d y$ , so $\frac{d y}{y ^{3}} = - t d t$ . Then $\int \frac{1}{y ^{3}} e^{- 1/ y} d y = - \int t e^{t} d t = - (t - 1) e^{t} + C = (1 - t) e^{t} + C$ .

Therefore $x e^{- 1/ y} = (1 + 1/ y) e^{- 1/ y} + C$ , i.e. $x = 1 + 1/ y + C e^{1/ y}$ .

Initial condition $x (1) = 1$ : $1 = 1 + 1 + C e \Rightarrow C = - 1/ e$ .

At $y = 1/2$ : $x = 1 + 2 + (- 1/ e) e^{2} = 3 - e$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q12: Methods of Solving First Order, First Degree Differential Equation

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