Let y=y(x) be the solution of the differential equation sec x d d +\ 2(1- x ) tan x + x (2- x )\ d x=0 such that y(0)=2 . Then y(2) is equal to :

Mathematics · Differential Equations

JEE Main 2024 — 30 January, Shift 1 — Question 9

Let $y=y(x)$ be the solution of the differential equation $\sec \mathrm{x} d \mathrm{~d}+\{2(1-\mathrm{x}) \tan \mathrm{x}+\mathrm{x}(2-\mathrm{x})\}$ $d x=0$ such that $y(0)=2$ . Then $y(2)$ is equal to :

Option A:
$2$
Correct
Option B:
$2\{1-\sin (2)\}$
Option C:
$2\{\sin (2)+1\}$
Option D:
$1$

Answer: A

Step-by-step solution

$\frac{d y}{d x}=2(x-1) \sin x+\left(x^{2}-2 x\right) \cos x$

Now both side integrate $y(x)=\int 2(x-1) \sin x d x+\left[\left(x^{2}-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right]$

$y(x)=\left(x^{2}-2 x\right) \sin x+\lambda$

$\mathrm{y}(0)=0+\lambda \Rightarrow 2=\lambda$

$y(x)=\left(x^{2}-2 x\right) \sin x+2$

$y(2)=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential

Let y = y ( x ) be the solution of the differential equation ( 1 - x^2 )dy = [ xy + ( x^3 + 2 )sqrt 3 ( 1 - x^2 ) ]dx , - 1 < x < 1,y ( 0 )…

Answer: A

Step-by-step solution

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