JEE Advanced 2023 — Paper 2 — MATH-III solution

JEE Advanced2023Paper 2MATH-III

For $x \in R$ , let $y (x)$ be a solution of the differential equation

$(x^{2} - 5) \frac{d y}{d x} - 2 x y = - 2 x (x^{2} - 5)^{2}$ such that $y (2) = 7$ .

Then the maximum value of the function $y (x)$ is

Your answer

Solution

Divide by $x^{2} - 5$ to obtain the linear form $\frac{d y}{d x} - \frac{2 x}{x ^{2} - 5} y = - 2 x (x^{2} - 5) .$

The integrating factor is $exp (- \int \frac{2 x}{x ^{2} - 5} d x) = \frac{1}{x ^{2} - 5}$ . Multiplying through,

$\frac{d}{d x} (\frac{y}{x ^{2} - 5}) = - 2 x \Rightarrow \frac{y}{x ^{2} - 5} = - x^{2} + C .$

Using $y (2) = 7$ : $\frac{7}{- 1} = - 4 + C$ , so $C = - 3$ . Therefore $y = - (x^{2} - 5) (x^{2} + 3)$ .

Let $t = x^{2} \geq 0$ . Then $y = - (t - 5) (t + 3) = - t^{2} + 2 t + 15$ , whose vertex is at $t = 1$ . So $y_{m a x} = - 1 + 2 + 15 = 16$ , attained at $x^{2} = 1$ .

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