JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0$ , $x > 2$ , passes through the point $(3, 15)$ , then the local maximum value of $f$ is _________.

Your answer

Solution

Rewrite: $(x^{2} - 4) y^{'} - 2 x y = - 2 x (4 - x^{2})^{2} = - 2 x (x^{2} - 4)^{2}$ .

Divide by $(x^{2} - 4)^{2}$ :

$\frac{y ^{'}}{x ^{2} - 4} - \frac{2 x y}{( x ^{2} - 4 ) ^{2}} = - 2 x$ .

The left side is $\frac{d}{d x} [\frac{y}{x ^{2} - 4}]$ , so $\frac{y}{x ^{2} - 4} = - x^{2} + C$ .

At $(3, 15)$ : $\frac{15}{5} = - 9 + C \Rightarrow C = 12$ . Hence $y = (12 - x^{2}) (x^{2} - 4)$ .

Differentiate: $y^{'} = - 2 x (x^{2} - 4) + (12 - x^{2}) (2 x) = 2 x [(12 - x^{2}) - (x^{2} - 4)] = 2 x (16 - 2 x^{2}) = 4 x (8 - x^{2})$ .

Critical point with $x > 2$ : $x = 22$ . Second derivative changes sign from + to -, so it is a local max.

$y (22) = (12 - 8) (8 - 4) = 4 \cdot 4 = 16$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →