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 \_\_\_\_ .

Correct answer: 16 Rewriting: (dy)/(dx) - (2x)/(x^2-4)y = -2x(x^2-4) . IF = e^ -∫ (2x)/(x^2-4)dx = (1)/(x^2-4) . (d)/(dx)((y)/(x^2-4)) = -2x (y)/(x^2-4) = -x^2 + C . y = (x^2-4)(C-x^2) . At (3,15) : 15 = 5(C-9) C = 12 . y = (x^2-4)(12-x^2) . y' = 4x(8-x^2) = 0 x = 2 2 (for x > 2 ). y'' = 32 - 12x^2 = -64 f(2 2 ) = (8-4)(12-8) = 16 .

JEE Main 202623 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Differential Equations

If the solution curve y=f(x) of the differential equation (x^ 2 -4)...

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

,<br> passes through the point

(3,15)

, then the local maximum value of

is

\_\_\_\_$.

Correct answer

Step-by-step 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

x = 2\sqrt{2}

, confirming local maximum.

f(2\sqrt{2}) = (8-4)(12-8) = 16

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