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

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Differential Equations

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

Answer: 16

Write the equation as

(x^{2} - 4) y^{'} - 2 x y = - 2 x (x^{2} - 4)^{2}

and divide by

(x^{2} - 4)^{2}

; the left side is

\frac{d}{d x} (\frac{y}{x ^{2} - 4}) = - 2 x

. Integrating,

\frac{y}{x ^{2} - 4} = - x^{2} + C

, so

y = (C - x^{2}) (x^{2} - 4)

. The point

(3, 15)

gives

15 = (C - 9) (5)

, so

C = 12

and

y = (12 - x^{2}) (x^{2} - 4)

. Then

y^{'} = 4 x (8 - x^{2})

, which for

x > 2

vanishes at

x = 22

and changes from positive to negative there. The local maximum value is

y (22) = (12 - 8) (8 - 4) = 16

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Integer answer

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