For x in mathbb R , let y(x) be a solution of the differential equation (x 2 -5 ) 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

Mathematics · Differential Equations

JEE Advanced 2023 — Paper 2 — Question 9

For $x \in \mathbb{R}$ , let $y(x)$ be a solution of the differential equation $\left(x^{2}-5\right) \frac{d y}{d x}-2 x y=-2 x\left(x^{2}-5\right)^{2}$ such that

$y(2)=7$ . Then the maximum value of the function $y(x)$ is

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$\left(x^{2}-5\right) \frac{d y}{d x}-2 x y=-2 x\left(x^{2}-5\right)^{2}$ $\frac{d y}{d x}-\frac{2 x}{x^{2}-5} y=-2 x\left(x^{2}-5\right)$ I.F. $=e^{\int-\frac{2 x}{x^{2}-5} d x}=e^{-\ln \left(x^{2}-5\right)}$ $=\frac{1}{x^{2}-5}$ $y \cdot \frac{1}{x^{2}-5}=\int-2 x d x$ $y \cdot \frac{1}{x^{2}-5}=-x^{2}+c$ $-7=-4+c \Rightarrow c=-3$ Now $y=-\left(x^{4}-2 x^{2}-15\right)$

y=-\left(\left(x^{2}-1\right)^{2}-16\right)

so maximum value $=16$

Answer key and solution verified before publishing.

Practise Differential Equations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential