JEE Main 2025 Jan 24 Shift 2, Mathematics Q4: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Jan 24, Shift 2Mathematics

Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f^{'} (x) = 2 x f (x) + 3$ , with $f (1) = 4$ . Then $2 f (2)$ is equal to:

A
$29$
B
$19$
C
$39$
D
$23$

Solution

Rearrange the ODE so the left side becomes the derivative of a quotient:

$x^{2} f^{'} (x) - 2 x f (x) = 3 \Rightarrow \frac{x ^{2} f ^{'} ( x ) - 2 x f ( x )}{x ^{4}} = \frac{3}{x ^{4}}$

The left side is exactly $\frac{d}{d x} (\frac{f ( x )}{x ^{2}})$ .

$\frac{d}{d x} (\frac{f ( x )}{x ^{2}}) = \frac{3}{x ^{4}}$

Integrate both sides:

$\frac{f ( x )}{x ^{2}} = - \frac{1}{x ^{3}} + C \Rightarrow f (x) = - \frac{1}{x} + C x^{2}$

Apply $f (1) = 4$ : $4 = - 1 + C \Rightarrow C = 5$ . So $f (x) = - \frac{1}{x} + 5 x^{2}$ .

$2 f (2) = 2 (- \frac{1}{2} + 5 \cdot 4) = 2 (- \frac{1}{2} + 20) = - 1 + 40 = 39$

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q4: Methods of Solving First Order, First Degree Differential Equation

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