Q46 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
If the solution curve $y=f(x)$ of the differential equation
$\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$,
passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
$\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$,
passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
Correct answer: 16
Solution
Rewriting: $\frac{dy}{dx} - \frac{2x}{x^2-4}y = -2x(x^2-4)$.
IF $= e^{-\int \frac{2x}{x^2-4}dx} = \frac{1}{x^2-4}$.
$\frac{d}{dx}\left(\frac{y}{x^2-4}\right) = -2x \Rightarrow \frac{y}{x^2-4} = -x^2 + C$.
$y = (x^2-4)(C-x^2)$. At $(3,15)$: $15 = 5(C-9) \Rightarrow C = 12$.
$y = (x^2-4)(12-x^2)$.
$y' = 4x(8-x^2) = 0 \Rightarrow x = 2\sqrt{2}$ (for $x > 2$).
$y'' = 32 - 12x^2 = -64 < 0$ at $x = 2\sqrt{2}$, confirming local maximum.
$f(2\sqrt{2}) = (8-4)(12-8) = 16$.
IF $= e^{-\int \frac{2x}{x^2-4}dx} = \frac{1}{x^2-4}$.
$\frac{d}{dx}\left(\frac{y}{x^2-4}\right) = -2x \Rightarrow \frac{y}{x^2-4} = -x^2 + C$.
$y = (x^2-4)(C-x^2)$. At $(3,15)$: $15 = 5(C-9) \Rightarrow C = 12$.
$y = (x^2-4)(12-x^2)$.
$y' = 4x(8-x^2) = 0 \Rightarrow x = 2\sqrt{2}$ (for $x > 2$).
$y'' = 32 - 12x^2 = -64 < 0$ at $x = 2\sqrt{2}$, confirming local maximum.
$f(2\sqrt{2}) = (8-4)(12-8) = 16$.
Hence, the correct answer is 16.