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Test: Single Correct MCQs: Differential Equations | JEE Advanced


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9 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Differential Equations | JEE Advanced

Test: Single Correct MCQs: Differential Equations | JEE Advanced for JEE 2022 is part of Mathematics For JEE preparation. The Test: Single Correct MCQs: Differential Equations | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: Single Correct MCQs: Differential Equations | JEE Advanced MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Single Correct MCQs: Differential Equations | JEE Advanced below.
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Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 1

A solution of the differential equation 

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 1

By actual verification we find that the choice (c), i.e. y = 2x-4 satisfies the given differential equation.

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 2

If x2 + y2 = 1, then

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 2

Given x2 + y2 = 1. Differen tiat ing w.r.t . x , we get 2x + 2yy ' = 0 or x + yy '= 0. Again differentiating w.r.t. x, we get 1 + y ' y '+ yy " = 0 or 1 + (y')2 + yy" = 0

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 3

If y(t) is a solution of  then y(1) is equal to

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 3

The given differential equation is




Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 4

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 4

Integrating both sides
⇒ ln(1 + y) = - ln(2 + sin x)+C
Put x = 0 andy=1 ⇒ ln(2) = - ln 2+ C ⇒ C = ln4

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 5

If y = y(x) and it follows the relation x cos y + y cos x = π then y" (0) =

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 5

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 6

The solution of primitive integral equation (x2 + y2) dy = xy dx is y = y(x). If y (1) = 1 and (x0) = e, then x0 is equal to

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 6

The given D.E. is (x2 + y2)dy = xy dx s.t. y (1) = 1 and y(x0) = e

The given eqn can be written as



Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 7

For th e primitive in tegral equation ydx + y2dy = x dy; x ∈ R, y > 0, y = y(x), y(1) = 1, then y(–3) is

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 7

The given eqn is



Now to find y(-3), puttin g x =-3 in above eqn, we get

But given that y > 0 , ∴ y = 3

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 8

The differential equation  determines a family of circles with

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 8


⇒ ( x - c)2 + y2 = 1

which is a circle of fixed radius 1 and variable centre (c, 0) lying on x-axis.

Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 9

The function y = f (x) is the solution of the differential equation  satisfying f (0) = 0. Then 

Detailed Solution for Test: Single Correct MCQs: Differential Equations | JEE Advanced - Question 9

Given D.E. can be written as






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