JEE Advanced 2023 · Paper 2 · Mathematics Q9

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MathematicsInteger+4Question 9 of 51

For $x \in \mathbb{R}$ , let $y(x)$ be a solution of the differential equation $(x^2-5)\dfrac{dy}{dx} - 2xy = -2x(x^2-5)^2$ such that $y(2) = 7$ . Then the maximum value of the function $y(x)$ is _____.

Your answer

Answer and worked solution

Official answer: 16

Exact derivative. Divide by $(x^2-5)^2$ :
$\frac{(x^2-5)y'-2xy}{(x^2-5)^2}=\frac{d}{dx}\left(\frac{y}{x^2-5}\right)=-2x.$

Integrate.
$\frac{y}{x^2-5}=-x^2+C.$
$y(2)=7$ gives $-7=-4+C$ , so $C=-3$ and
$y=-(x^2-5)(x^2+3)=-x^4+2x^2+15.$

Maximum. $y'=-4x^3+4x=0$ at $x=0,\pm1$ . Then $y(\pm1)=16$ and $y(0)=15$ , so the maximum is $\mathbf{16}$ .

Topics:Linear first-order equations Maxima and minima

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JEE Advanced 2023 · Paper 2, Mathematics question 9