JEE Main 2026 Apr 6 Shift 2, Mathematics Q18: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 6, Shift 2Mathematics

Let $f : R \to R$ be such that $f (x y) = f (x) f (y)$ , for all $x, y \in R$ and $f (0) \neq = 0$ . Let $g : [1, \infty) \to R$ be a differentiable function such that $x^{2} g (x) = \int_{1}^{x} (t^{2} f (t) - t g (t)) d t$ . Then $g (2)$ is equal to:

A
$\frac{13}{8}$
B
$\frac{11}{16}$
C
$\frac{15}{32}$
D
$\frac{17}{64}$

Solution

Identify $f$ . Setting $y = 0$ : $f (0) = f (x) f (0)$ . Since $f (0) \neq = 0$ , $f (x) = 1$ for all $x$ .

Differentiate the integral relation using the fundamental theorem:

$\frac{d}{d x} [x^{2} g (x)] = x^{2} f (x) - xg (x)$

$2 xg (x) + x^{2} g^{'} (x) = x^{2} - xg (x)$

$x^{2} g^{'} (x) + 3 xg (x) = x^{2}$ $\Rightarrow$ $g^{'} (x) + \frac{3}{x} g (x) = 1$ .

Solve the linear ODE. Integrating factor $= e^{3 l n x} = x^{3}$ :

$\frac{d}{d x} [x^{3} g (x)] = x^{3}$ $\Rightarrow$ $x^{3} g (x) = \frac{x ^{4}}{4} + C$ .

The integral relation gives $g (1) = 0$ (right side is 0 when $x = 1$ ), so $C = - \frac{1}{4}$ .

At $x = 2$ : $8 g (2) = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$ $\Rightarrow$ $g (2) = \frac{15}{32}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q18: Methods of Solving First Order, First Degree Differential Equation

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