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Differential Equations - 1 - Mathematics MCQ


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Differential Equations - 1

Differential Equations - 1 for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation. The Differential Equations - 1 questions and answers have been prepared according to the Mathematics exam syllabus.The Differential Equations - 1 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Differential Equations - 1 below.
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Differential Equations - 1 - Question 1

Let k be real constant. The solution of the differential equations  satisfies the relation

Detailed Solution for Differential Equations - 1 - Question 1

we have dy/dx = 2y + z and ...(i)
dz/dx = 3y  ...(ii)
from 3 (i) l (ii), we get 

implies 
on integrating, we get

implies ln(3y + z) = 3x + c
So, 3y + z = ke3x

Differential Equations - 1 - Question 2

Find the General Solution of the given differential equation.

Detailed Solution for Differential Equations - 1 - Question 2

This special case where D has a co-efficient is solved using Legendre’s method. 

Let log⁡(8x + 7) = z 

Then, ez = 8x + 7

Substituting this in the equation, 

8Dy + 2y = x 

y(8D + 2) = x 

Thus the auxiliary equation is 8D + 2 = 0 

Thus, D = -1 / 4

(Solving by substituting powers of ez)

The General solution is C.F + P.I =

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Differential Equations - 1 - Question 3

The differential equation 2ydx - (3y - 2x)dy = 0

Detailed Solution for Differential Equations - 1 - Question 3

Here

Differential Equations - 1 - Question 4

Consider the differential equation 2 cos (y2)dx - xy sin (y2)dy = 0

Detailed Solution for Differential Equations - 1 - Question 4

Correct Answer :- D

Explanation : 2cos (y2)dx - xy sin (y2) dy = 0

Multiplying by x3

2x3 cos(y2)dx - x4y sin(y2)dy = 0

dM/dy = - 4x3y sin(y2)

dN/dx = - 4x3y sin(y2)

Since dM/dy = dN/dx

So, x3 is an integrating factor.

Differential Equations - 1 - Question 5

If the general solutions of a differential equation are (y + c)2 = cx, where c is an arbitrary constant, then the order and degree of differential equation is  

Detailed Solution for Differential Equations - 1 - Question 5

There will only one constant in the first-order differential equation. Differentiating the given equation.

 

Putting the value of c in Eq. (1) and simplifying we will get a first-order and second-degree equation. Hence, (A) is the correct answer.

Differential Equations - 1 - Question 6

An integration factor of  is

Detailed Solution for Differential Equations - 1 - Question 6

we have 

Differential Equations - 1 - Question 7

The general solution of 
(Here c1 and c2 are arbitrary constants.)

Detailed Solution for Differential Equations - 1 - Question 7

Let z = log x, Then

The differential equation changes to

implies

So, 

Differential Equations - 1 - Question 8

Solution of the differential equation

Detailed Solution for Differential Equations - 1 - Question 8


=> x2 ln⁡x dy - xy dy=xy dx – y2 ln⁡y dx …….dividing by x2 y2 then


on integrating we get

where c is a constant of integration.

Differential Equations - 1 - Question 9

Let y1(x) and y2(x) be linearly independent solution of the differential equation y" + P(x) y' + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) dy2(x) are linearly independent solutions of the given differential equations if

Detailed Solution for Differential Equations - 1 - Question 9

y3(x) and y4(x) are linearly dependent if y3(x) = ky4(x) where k is any constant 

Since, y1(x) and y2(x) are independent
So, a - kc = 0
and b - kd = 0
or 
So, y3(x) and y4(x) are independent if

implies ad ≠ bc

Differential Equations - 1 - Question 10

Consider the differential equations, satisfying y(0) = 0,  where  This initial value problem

Detailed Solution for Differential Equations - 1 - Question 10

we have 
or, 
Thus, 
implies y = x2 + c for x ≤ 0,
but y(0) = 0 implies c = 0
So,
 

Differential Equations - 1 - Question 11

The particular integral of the differential equation 
y" + y' + 3y = 5 cos(2x + 3) is

Detailed Solution for Differential Equations - 1 - Question 11

we have (D2 + D + 3)y = 5 cos(2x + 3)
So,

Differential Equations - 1 - Question 12

The general solution of differential equation
4x2y" - 8xy' + 9y = 0 is 

Detailed Solution for Differential Equations - 1 - Question 12

Let z = log x, Then 

So, The differential equation becomes

So, 

Differential Equations - 1 - Question 13

What is the order of the non-homogeneous partial differential equation,

Detailed Solution for Differential Equations - 1 - Question 13

The order of an equation is defined as the highest order derivative present in the equation and hence from the equation,it is clear that it of 2nd order.

Differential Equations - 1 - Question 14

If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation is

Detailed Solution for Differential Equations - 1 - Question 14


Differential Equations - 1 - Question 15

A general solution of the differential equation 

Detailed Solution for Differential Equations - 1 - Question 15

Differential Equations - 1 - Question 16

Consider the differential equation  Then   is equal to

Detailed Solution for Differential Equations - 1 - Question 16

we have 
implies 

Differential Equations - 1 - Question 17

Consider the differential equation dy/dx = ay - by2,  where a, b > 0 and y(0) = y0.  As  x → + ∝, the solution y (x) tends to

Detailed Solution for Differential Equations - 1 - Question 17


So, Equation becomes,


Differential Equations - 1 - Question 18

Orthogonal trajectories of the family of curves (x - 1)2 + y2 + 2ax = 0 are the solution of the differential equation

Detailed Solution for Differential Equations - 1 - Question 18



By putting the value of a from equation (ii) in equation (i), we get.


It ’s orthogonal trajectory will be D.E.

Differential Equations - 1 - Question 19

The solution of differential equation  represents the family of 

Detailed Solution for Differential Equations - 1 - Question 19

Given equation, 

Dividing both sides by 2x, we get 

There is a linear differential equation of the form: 

The integrating factor is found by:

Multiply both sides of the equation by the integrating factor:

The solution  represents a family of parabolas. This is because the equation has the form of a quadratic curve (which describes a parabola). 

Therefore, the answer is c) parabolas

Differential Equations - 1 - Question 20

Let  then sum of degree and order of given ordinary differential equation is 

Detailed Solution for Differential Equations - 1 - Question 20

Given, 

Taking the natural logarithm of both sides to eliminate the exponential term:

The order of a differential equation is the highest derivative present in the equation.

Here, the highest derivative is y′′y''y′′ (second derivative), so the order of the equation is 2.

The degree of a differential equation is the exponent of the highest derivative, provided that the equation is polynomial in the derivatives.

In this case, the equation llny=y′+y′′ is not polynomial in y′ and y′′ due to the presence of the logarithmic function. Therefore, the degree is 1.

The order is 2, and the degree is 1.

Thus, the sum of the degree and order is 2+1=3

(a) is the correct answer.

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