Differential Equations – Solution of Differential Equations – JEE Main 23 Jan 2026 Shift 2

Solution:

Rearrange the differential equation:
$$ (x^2 – 4)\frac{dy}{dx} – 2xy = -2x(4 – x^2)^2 $$

Divide by $(x^2 – 4)^2$:
$$ \frac{(x^2 – 4)dy – 2xy dx}{(x^2 – 4)^2} = \frac{-2x(4 – x^2)^2}{(x^2 – 4)^2} dx $$

$$ d\left(\frac{y}{x^2 – 4}\right) = -2x dx $$

Integrating both sides:
$$ \frac{y}{x^2 – 4} = -x^2 + C $$
$$ y = (x^2 – 4)(C – x^2) $$

Given that the curve passes through (3, 15):
$$ 15 = (3^2 – 4)(C – 3^2) $$
$$ 15 = 5(C – 9) $$
$$ 3 = C – 9 \Rightarrow C = 12 $$

So, the function is:
$$ y = (x^2 – 4)(12 – x^2) $$

To find the local maximum, find $y’$:
$$ y’ = 2x(12 – x^2) + (x^2 – 4)(-2x) $$
$$ y’ = 24x – 2x^3 – 2x^3 + 8x $$
$$ y’ = 32x – 4x^3 $$

Set $y’ = 0$:
$$ 4x(8 – x^2) = 0 $$
Since $x > 2$, $x = \sqrt{8} = 2\sqrt{2}$.

The local maximum value is $f(2\sqrt{2})$:
$$ y = ((2\sqrt{2})^2 – 4)(12 – (2\sqrt{2})^2) $$
$$ y = (8 – 4)(12 – 8) = 4 \times 4 = 16 $$

Ans. (16)