JEE Advanced 2023 Paper 2 · Q09 · Differential Equations

Question

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 ___.

SolveKar AI · Accepted Answer

Rewrite as $\dfrac{dy}{dx} - \dfrac{2x}{x^2-5}y = -2x(x^2-5)$. Integrating factor: $e^{-\int 2x/(x^2-5)\,dx} = 1/|x^2-5|$. Solution (for $x^2 < 5$): $y/(5-x^2) = \int 2x\,dx + C = x^2 + C$, so $y = (x^2+C)(5-x^2)$. Using $y(2) = 7$: $(4+C)(1) = 7 \Rightarrow C = 3$. Therefore $y = (x^2+3)(5-x^2) = -x^4 + 2x^2 + 15$. $dy/dx = -4x^3 + 4x = -4x(x^2-1) = 0 \Rightarrow x = 0, \pm 1$. At $x = \pm 1$: $y = 4\cdot 4 = 16$ (maximum).