Let y=y(x) be the solution of the differential equation (x 2 - x√(x 2-1))dy + (y(x-√(x 2-1)) - x)dx = 0 , x≥1. If y(1)=1, then the greatest integer less than y(√5) is .

Mathematics · Differential Equations

JEE Main 2026 — 6 April, Morning Shift — Question 45

Let y=y(x) be the solution of the differential equation $(x^2 - x\sqrt{x^2-1})dy + (y(x-\sqrt{x^2-1}) - x)dx = 0$ , x≥1. If y(1)=1, then the greatest integer less than y(√5) is ______.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\frac{d y}{d x}+\frac{y}{x}=\frac{1}{x-\sqrt{x^{2}-1}}$ $\frac{d y}{d x}+\frac{y}{x}=x+\sqrt{x^{2}-1}$ I. $f=e^{\ln x}=x$ $y \cdot x=\int\left(x+\sqrt{x^{2}-1}\right) x d x$ $=\frac{\mathrm{x}^{2}}{2}+\frac{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}{3}+\mathrm{C}$ Given $y(1)=1$ $1=\frac{1}{2}+\mathrm{C} \Rightarrow \mathrm{C}=\frac{1}{2}$ $[\mathrm{y}(\sqrt{5})]=\left[\frac{\sqrt{5}}{2}+\frac{8}{3}+\frac{1}{2 \sqrt{5}}\right]=3$

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential