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 _____. [2026]

Question

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 _____. [2026]

Smart Achievers · Accepted Answer

(16)

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

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

$\Rightarrow y = (- x^{2} + C) (x^{2} - 4)$

$for x = 3, y = 15 \Rightarrow C = 12$

$\Rightarrow y = (- x^{2} + 12) (x^{2} - 4)$

$y' = 0 \Rightarrow x = 2 \sqrt{2}$

$y_{local max} = ({(2 \sqrt{2})}^{2} - 4) (- {(2 \sqrt{2})}^{2} + 12)$

$= 16$

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