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Study Stuff for MCA Examinations - NIMCET
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Previous Year Question (PYQs)
If the solution curve $y = f(x)$ of the differential equation
$(x^2 - 4)y' - 2xy + 2x(4 - x)^2 = 0,; x > 2$,
passes through the point $(3, 15)$, then the local maximum value of $f$ is ______.
Solution
$(x^2 - 4)y' - 2xy = -2x(x^2 - 4)^2$
$\dfrac{d}{dx}\left(\dfrac{y}{x^2 - 4}\right) = -2x$
$y = (-x^2 + C)(x^2 - 4)$
For $x = 3,; y = 15$
$15 = (-9 + C)(5) \Rightarrow C = 12$
$y = (-x^2 + 12)(x^2 - 4)$
For maxima, $y' = 0 \Rightarrow x = 2\sqrt{2}$
$y_{\text{max}} = (-(2\sqrt{2})^2 + 12)\big((2\sqrt{2})^2 - 4\big)$
$= ( -8 + 12)(8 - 4) = 4 \cdot 4 = 16$
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