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

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Differential Equations

For

x \in R

, let

y (x)

be a solution of the differential equation

(x^{2} - 5) \frac{dy}{dx} - 2 xy = - 2 x (x^{2} - 5)^{2}

such that

y (2) = 7

.
Then the maximum value of the function

y (x)

Answer: 16

Rewrite as:

\frac{d y}{d x} - \frac{2 x}{x ^{2} - 5} y = - 2 x (x^{2} - 5)

Integrating factor:

I F = e^{- \int \frac{2 x}{x ^{2} - 5} d x} = e^{- l n (x^{2} - 5)} = \frac{1}{x ^{2} - 5}

Multiplying through:

\frac{y}{x ^{2} - 5} = \int - 2 x d x = - x^{2} + C

y = (x^{2} - 5) (- x^{2} + C)

Using

y (2) = 7

7 = (4 - 5) (- 4 + C) = - (C - 4) = 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) = - t^{2} + 2 t + 15

This is a downward parabola in

t

with maximum at

t = 1

y_{m a x} = - 1 + 2 + 15 = 16

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer