JEE Main 2026 Apr 2 Shift 2, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 2, Shift 2Mathematics

Let $x = x (y)$ be the solution of the differential equation $2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0$ , $y > 1$ , $x (e) = e$ . Then $x (e^{2})$ is equal to:

A
$\frac{3}{2} e^{2}$
B
$\frac{2}{3} e^{2}$
C
$e^{2}$
D
$2 e^{2}$

Solution

Rewrite as $2 y (y d x - x d y) + x^{2} d y = 0$ . Note $d (\frac{y}{x}) = \frac{x d y - y d x}{x ^{2}}$ , so $y d x - x d y = - x^{2} d (\frac{y}{x})$ .

Substituting: $- 2 y x^{2} d (\frac{y}{x}) + x^{2} d y = 0$ , i.e. $- 2 y d (\frac{y}{x}) + d y = 0$ .

Divide by $y$ : $- 2 d (\frac{y}{x}) + \frac{d y}{y} = 0$ .

Integrating: $- \frac{2 y}{x} + ln y = C$ .

Applying $x (e) = e$ : $- 2 + 1 = C \Rightarrow C = - 1$ . The solution is $\frac{2 y}{x} - ln y = 1$ .

At $y = e^{2}$ : $\frac{2 e ^{2}}{x} - 2 = 1$ , so $\frac{2 e ^{2}}{x} = 3$ , giving $x = \frac{2 e ^{2}}{3}$ .

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JEE Main 2026 Apr 2 Shift 2, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

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