Which of the following functions is a solution to the differential equation $$y'' - 4y' + 4y = 0$$?

Correct Option: A. Solution: The characteristic equation of the differential equation $$y'' - 4y' + 4y = 0$$ is $$r^2 - 4r + 4 = 0$$, which factors to $$(r - 2)^2 = 0$$. This has a repeated root at $$r = 2$$, so the general solution is $$y = (C_1 + C_2 x)e^{2x}$$. Hence, $$y = e^{2x}$$ is a particular solution.

At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$. Find the equation of the curve given that it passes through $(-2, 1)$.

Correct Option: B. Solution: The slope of the tangent at $(x, y)$ is given by $\frac{dy}{dx}$. The slope of the line segment joining $(x, y)$ to $(-4, -3)$ is $\frac{y + 3}{x + 4}$. Thus, $\frac{dy}{dx} = 2 \cdot \frac{y + 3}{x + 4}$. Solving this differential equation with the initial condition $(-2, 1)$, we find the equation of the curve to be $y = x^2 + 3x + 1$.

Which of the following is the general solution of the differential equation $$\frac{dy}{dx} = 0$$?

Correct Option: A. Solution: If $$\frac{dy}{dx} = 0$$, then $$y$$ is a constant function. Thus, the general solution is $$y = C$$.

Differential Equations

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Summary

Chapter 9: Differential Equations

Summary

A differential equation involves derivatives of a dependent variable with respect to independent variables.
The order of a differential equation is determined by the highest order derivative present.
The degree of a differential equation is defined if it is a polynomial equation in its derivatives.
A solution to a differential equation is a function that satisfies the equation when substituted.
General solutions contain arbitrary constants equal to the order of the differential equation; particular solutions do not.
The variable separable method is used for equations where variables can be completely separated.
Homogeneous differential equations can be expressed in a specific form involving homogeneous functions of degree zero.
A first-order linear differential equation has the form Py = Q, where P and Q are constants or functions of x only.

Learning Objectives

Understand the definition of differential equations.
Identify the order and degree of differential equations.
Distinguish between general and particular solutions of differential equations.
Apply methods to solve first-order differential equations.
Recognize the importance of differential equations in various scientific fields.
Solve homogeneous differential equations using appropriate substitutions.
Utilize the method of separation of variables for solving differential equations.
Analyze the applications of differential equations in real-world scenarios.

Detailed Notes

Chapter 9: Differential Equations

9.1 Introduction

Definition: A differential equation is an equation involving derivatives of a dependent variable with respect to independent variable(s).
Importance: Differential equations arise in various fields such as Physics, Chemistry, Biology, and Economics.
Objective: This chapter covers basic concepts, general and particular solutions, formation, methods to solve first order - first degree differential equations, and applications.

9.2 Basic Concepts

9.2.1 Order of a Differential Equation

Definition: The order of a differential equation is the order of the highest derivative present.
Examples:
- Equation (6): Order 1
- Equation (9): Order 2
- Equation (10): Order 1
- Equation (11): Order not defined

9.2.2 Degree of a Differential Equation

Definition: The degree is defined if the differential equation is a polynomial in its derivatives.
Examples:
- Equation (9): Degree 1
- Equation (10): Degree 2
- Equation (11): Degree not defined

9.3 General and Particular Solutions of a Differential Equation

General Solution: Contains arbitrary constants equal to the order of the differential equation.
Particular Solution: Free from arbitrary constants.
Example: For the equation dy/dx = g(x), the solution curve is y = Φ(x).

9.4 Methods of Solving Differential Equations

Variable Separable Method: Used when variables can be separated completely.
Homogeneous Differential Equations: Can be expressed in the form P(y) = Q(x).

9.5 Applications of Differential Equations

Example: The volume of a spherical balloon changes at a constant rate. If the radius is initially 3 units and after 3 seconds it is 6 units, find the radius after t seconds.

Exercises

Find a particular solution for the differential equation dy/dx = y tan x; y = 1 when x = 0.
Verify that the function y = c₁ e^(ax) cos(bx) is a solution of the differential equation.
Find the general solution of the differential equation dy/dx + x²y² + y + 1 = 0.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Differential Equations

Common Pitfalls

Misunderstanding the Definition: Students often confuse differential equations with algebraic equations. Remember, a differential equation involves derivatives of a dependent variable with respect to an independent variable.
Ignoring Initial Conditions: Failing to apply initial conditions when finding particular solutions can lead to incorrect answers.
Incorrectly Identifying Homogeneous Equations: Not all equations that appear to be homogeneous are so. Ensure to verify the definition of homogeneous functions.
Forgetting to Separate Variables: In separable differential equations, students sometimes forget to separate variables correctly before integrating.

Tips for Success

Understand the Concepts: Make sure you grasp the basic concepts of differential equations, including general and particular solutions.
Practice with Examples: Work through examples systematically to familiarize yourself with different types of differential equations and their solutions.
Check Your Work: After solving a differential equation, substitute your solution back into the original equation to verify correctness.
Use Graphs: Visualizing the solution curves can help in understanding the behavior of solutions to differential equations.
Review Common Forms: Familiarize yourself with common forms of differential equations and their respective solution methods.

Practice & Assessment

Multiple Choice Questions

r(t) = \sqrt{t + 4}

r(t) = \sqrt{2t + 4}

r(t) = \sqrt{3t + 4}

r(t) = \sqrt{4t + 4}

Correct Answer: B

Solution:

The surface area

S

of a sphere is given by

S = 4\pi r^2

. The rate of change of surface area is constant, so

\frac{dS}{dt} = k

. Given that

r(0) = 2

and

r(4) = 4

, we set up the equation

8\pi r \frac{dr}{dt} = k

. Solving this differential equation with the given conditions, we find

r(t) = \sqrt{2t + 4}

Correct Answer: C

Solution:

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

\frac{d^3y}{dx^3}

, so the order is 3.

(x^2 + xy) dy = (x^2 + y^2) dx

y dx + x dy = 0

y' + y = x

dy/dx = x + y

Correct Answer: A

Solution:

A differential equation is homogeneous if it can be expressed in the form

M(x, y) dy = N(x, y) dx

where both

M

and

N

are homogeneous functions of the same degree. The equation

(x^2 + xy) dy = (x^2 + y^2) dx

satisfies this condition.

y = \sec(x) + C \cos(x)

y = \cos(x) + C \sec(x)

y = \sin(x) + C \cos(x)

y = \tan(x) + C \sec(x)

Correct Answer: A

Solution:

This is a first-order linear differential equation. The integrating factor is

e^{\int \tan(x) \, dx} = \sec(x)

. Multiplying through by the integrating factor, we get

\frac{d}{dx}(y \sec(x)) = \sec^3(x)

. Integrating both sides gives

y \sec(x) = \int \sec^3(x) \, dx + C

. Solving this integral leads to

y = \sec(x) + C \cos(x)

\frac{d^2y}{dx^2} + 3y = 0

\frac{dy}{dx} + y = 0

y'' + y' + y = 0

y = mx + c

Correct Answer: A

Solution:

A second-order differential equation involves the second derivative of the function. Option A involves

\frac{d^2y}{dx^2}

, which is a second derivative.

y = -e^x \cos(x) + C

y = e^x \cos(x) + C

y = e^x \sin(x) + C

y = -e^x \sin(x) + C

Correct Answer: A

Solution:

Integrating the differential equation

y' = e^x \sin(x)

, we use integration by parts. Let

u = \sin(x)

and

dv = e^x dx

, then

du = \cos(x) dx

and

v = e^x

. The integral becomes

\int e^x \sin(x) dx = -e^x \cos(x) + C

. Thus, the solution passing through (0, 0) is

y = -e^x \cos(x) + C

. Since the curve passes through (0, 0),

C = 1

y'' + y = 0

y' + y = x

y''' + 3y'' = 2

y^{(4)} - y = 0

Correct Answer: B

Solution:

A first-order differential equation involves only the first derivative. Option (b) is the only equation that involves

y'

and no higher derivatives.

James Bernoulli

Joseph Louis Lagrange

Jules Henri Poincare

John Bernoulli

Correct Answer: C

Solution:

Jules Henri Poincare strongly advocated the use of the word 'solution' for differential equations, which replaced the earlier term 'integral'.

Not defined

Correct Answer: C

Solution:

The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. Here, the highest order derivative is

\frac{d^2y}{dx^2}

and its highest power is 3.

0.25 mol/L

0.33 mol/L

0.20 mol/L

0.10 mol/L

Correct Answer: A

Solution:

The rate of reaction can be modeled by the differential equation

\frac{dC}{dt} = -kC^2

. Separating variables and integrating, we get

\frac{1}{C} = kt + C_0

. Using the initial condition

C(0) = 1

, we find

C_0 = 1

. After 2 hours,

C = 0.5

, so

\frac{1}{0.5} = 2k + 1

. Solving for

k

, we find

k = 0.5

. After 4 hours,

\frac{1}{C} = 2 \times 0.5 \times 4 + 1 = 5

. Thus,

C = 0.2

mol/L.

y = x^3 + 1

y = x^3 + C

y = x^3 + 2

y = 3x^2 + 1

Correct Answer: A

Solution:

Integrating

\frac{dy}{dx} = 3x^2

gives

y = x^3 + C

. Using the condition

y = 1

when

x = 0

, we find

C = 1

. Thus, the particular solution is

y = x^3 + 1

Order: 2, Degree: 3

Order: 3, Degree: 2

Order: 2, Degree: 2

Order: 1, Degree: 3

Correct Answer: A

Solution:

The highest derivative is

\frac{d^2y}{dx^2}

, so the order is 2. The degree is the highest power of the highest order derivative, which is 3.

Not defined

Correct Answer: C

Solution:

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

\frac{d^3y}{dx^3}

, so the order is 3.

y = C \sec x

y = C \cos x

y = C \tan x

y = C \sin x

Correct Answer: A

Solution:

The differential equation is separable. Integrating both sides gives

\ln |y| = \ln |\sec x| + C

, leading to

y = C \sec x

y = x^3 + C

y = 3x^3 + C

y = x^3

y = 3x^2 + C

Correct Answer: A

Solution:

Integrating both sides with respect to

x

, we get

y = \int 3x^2 \, dx = x^3 + C

, where

C

is the constant of integration.

y = e^{\sin(x)}

y = e^{-\sin(x)}

y = \cos(x)

y = \sin(x)

Correct Answer: A

Solution:

The differential equation is separable. Separating variables, we have

\frac{dy}{y} = \tan(x) dx

. Integrating both sides,

\ln|y| = -\ln|\cos(x)| + C

. Solving for

y

, we get

y = e^{C} e^{\sin(x)}

. Given

y = 1

when

x = 0

, we find

C = 0

. Thus, the particular solution is

y = e^{\sin(x)}

y'' + y' + y = 0

y' = \sin(x)

y'' + \frac{1}{y} = 0

y' = e^x

Correct Answer: A

Solution:

A polynomial differential equation is one where the derivatives are in polynomial form. Option (a) is a polynomial in terms of

y'', y'

, and

y

y = Cx

x = Cy

y = x + C

x = y + C

Correct Answer: A

Solution:

Rewriting the differential equation

y dx - x dy = 0

gives

\frac{dy}{dx} = \frac{y}{x}

. This is a separable equation, and separating variables gives

\frac{dy}{y} = \frac{dx}{x}

. Integrating both sides yields

\ln|y| = \ln|x| + C

, or

y = Cx

. Thus, the general solution is

y = Cx

The solution is always a polynomial.

The solution can be an integral of the differential equation.

The solution is always a constant.

The solution is always a linear function.

Correct Answer: B

Solution:

In the context of differential equations, the term 'solution' can refer to an integral of the differential equation, as historically referred to by James Bernoulli.

5 units

6 units

7 units

8 units

Correct Answer: B

Solution:

The volume of a sphere is given by

V = \frac{4}{3} \pi r^3

. The rate of change of volume is

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 10

. Solving this differential equation with the initial condition

r(0) = 3

, we find

r(t) = 6

after 5 seconds.

y = e^{2x}

y = e^{-2x}

y = e^{4x}

y = e^{x}

Correct Answer: A

Solution:

The characteristic equation of the differential equation

y'' - 4y' + 4y = 0

r^2 - 4r + 4 = 0

, which factors to

(r - 2)^2 = 0

. This has a repeated root at

r = 2

, so the general solution is

y = (C_1 + C_2 x)e^{2x}

. Hence,

y = e^{2x}

is a particular solution.

y = e^{\sin x}

y = e^{\cos x}

y = e^{x}

y = e^{-\sin x}

Correct Answer: A

Solution:

The differential equation

dy/dx = y \tan x

can be solved by separation of variables. Integrating both sides gives

\ln y = \sin x + C

. Using the initial condition

y = 1

when

x = 0

, we find

C = 0

. Thus,

y = e^{\sin x}

is the particular solution.

Correct Answer: C

Solution:

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

y'''

, which is the third derivative, so the order is 3.

\frac{d^2y}{dx^2} + y = 0

(\frac{dy}{dx})^2 + y = 0

\frac{d^3y}{dx^3} + (\frac{dy}{dx})^2 = 0

\frac{d^2y}{dx^2} + (\frac{dy}{dx})^3 = 0

Correct Answer: D

Solution:

The degree of a differential equation is defined as the highest power of the highest order derivative, provided the equation is a polynomial in derivatives. Option D,

\frac{d^2y}{dx^2} + (\frac{dy}{dx})^3 = 0

, has the highest order derivative as

\frac{dy}{dx}

raised to the power of 3, making its degree 3.

y = x^2 - 2

y = x^2 + 2

y = x^2 - 4

y = x^2 + 4

Correct Answer: A

Solution:

The differential equation can be rewritten as

\frac{dy}{dx} - \frac{y}{x} = x

. This is a first-order linear differential equation. The integrating factor is

e^{-\int \frac{1}{x} \, dx} = \frac{1}{x}

. Multiplying through by the integrating factor, we get

\frac{d}{dx}(\frac{y}{x}) = 1

. Integrating both sides gives

\frac{y}{x} = x + C

. Solving for

y

, we get

y = x^2 + Cx

. Using the initial condition

(0, -2)

, we find

C = -2

. Thus, the equation of the curve is

y = x^2 - 2

y = x^2 + 2

y = x^2 + 1

y = x^2 + 3

y = x^2 + 4

Correct Answer: A

Solution:

Integrating

\frac{dy}{dx} = 2x

gives

y = x^2 + C

. Using the condition

y = 3

when

x = 1

, we find

C = 2

, so the solution is

y = x^2 + 2

r = 3 + t

r = 3t

r = 3 + 2t

r = 6t

Correct Answer: C

Solution:

The radius increases linearly with time. Since it goes from 3 to 6 in 3 seconds, the rate of change is 1 unit per second. Thus,

r = 3 + t

y = 2\sin(x)

y = 2\csc(x)

y = 2\tan(x)

y = 2\sec(x)

Correct Answer: A

Solution:

The differential equation

\frac{dy}{dx} = y \cot(x)

can be solved by separating variables:

\frac{1}{y} dy = \cot(x) dx

. Integrating both sides, we get

\ln|y| = \ln|\sin(x)| + C

, or

y = C \sin(x)

. Given that

y = 2

when

x = \frac{\pi}{4}

, we find

C = 2\sqrt{2}

. Thus, the particular solution is

y = 2\sin(x)

y = x + 1

y = x - 1

y = x

y = -x

Correct Answer: C

Solution:

Using the substitution

x-y = t

, we get

\frac{dy}{dx} = \frac{t}{2x}

. Solving this differential equation gives

y = x

. Since the solution passes through (1, 1), this is the particular solution.

Not defined

Correct Answer: C

Solution:

The degree is the highest power of the highest order derivative when the equation is a polynomial in derivatives. Here, the degree is 3.

y = vx

, where

v

is a function of

x

x = vy

, where

v

is a function of

y

y = x + v

, where

v

is a function of

x

x = y + v

, where

v

is a function of

y

Correct Answer: A

Solution:

The substitution

y = vx

transforms the equation into

\frac{d(vx)}{dx} = \frac{x^2 + (vx)^2}{x(vx)}

, which simplifies to a separable form. This substitution leverages the homogeneity of the differential equation.

y = x^2 + 2

y = \frac{x^2}{2} + 2

y = x^2 - 2

y = \frac{x^2}{2} - 2

Correct Answer: B

Solution:

Integrating

\frac{dy}{dx} = x

gives

y = \frac{x^2}{2} + C

. Using the condition

y = 2

when

x = 0

, we find

C = 2

. Thus, the particular solution is

y = \frac{x^2}{2} + 2

e^{\int \cot(x) dx}

e^{\int \tan(x) dx}

e^{\int \sin(x) dx}

e^{\int \cos(x) dx}

Correct Answer: A

Solution:

The given differential equation is in the form

\frac{dy}{dx} + P(x)y = Q(x)

, where

P(x) = \cot(x)

. The integrating factor is

e^{\int P(x) dx} = e^{\int \cot(x) dx} = e^{\ln|\sin(x)|} = \sin(x)

r(t) = 3 + t

r(t) = 3 + 2t

r(t) = 3t

r(t) = 6 + t

Correct Answer: A

Solution:

The volume

V

of a sphere is given by

V = \frac{4}{3} \pi r^3

. The rate of change of volume is constant, implying

\frac{dV}{dt} = k

. Given

r(0) = 3

and

r(3) = 6

, we find that the radius increases linearly with time:

r(t) = 3 + t

Not defined

Correct Answer: C

Solution:

The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. Here, the highest order derivative is

\frac{d^2y}{dx^2}

, and its highest power is 3.

y = Ce^{-\frac{x^3}{3}}

y = Ce^{\frac{x^3}{3}}

y = Cx^3

y = Cx^{-3}

Correct Answer: A

Solution:

This is a first-order linear differential equation. The integrating factor is

e^{\int x^2 dx} = e^{\frac{x^3}{3}}

. Multiplying through by the integrating factor, the left side becomes the derivative of

ye^{\frac{x^3}{3}}

. Integrating both sides gives

y = Ce^{-\frac{x^3}{3}}

y = A \cos(2x) + B \sin(2x)

y = Ae^{2x} + Be^{-2x}

y = A \cos(x) + B \sin(x)

y = Ae^{x} + Be^{-x}

Correct Answer: A

Solution:

The characteristic equation for the differential equation is

r^2 + 4 = 0

, which has roots

r = \pm 2i

. The general solution for such a differential equation is

y = A \cos(2x) + B \sin(2x)

0.5 mol/L

0.25 mol/L

1 mol/L

0 mol/L

Correct Answer: B

Solution:

The rate of reaction can be modeled by the differential equation

\frac{dC}{dt} = -kC

, where

C

is the concentration. Solving this gives

C(t) = C_0 e^{-kt}

. Given

C(3) = 1

mol/L and

C(0) = 2

mol/L, we find

k = \frac{\ln(2)}{3}

. Thus,

C(6) = 2 e^{-2\ln(2)} = 0.25

mol/L.

y'' + y' + y = 0

y' + \sin(y) = 0

y''' + y''^2 + y = 0

y'' + y^2 = 0

Correct Answer: B

Solution:

The differential equation

y' + \sin(y) = 0

is not a polynomial in its derivatives, hence its degree is not defined. Degree is defined only for polynomial differential equations.

y = 2x + 5

y = x^2 + 3x + 1

y = 2x^2 + 3x + 1

y = 3x^2 + 4x + 1

Correct Answer: B

Solution:

The slope of the tangent at

(x, y)

is given by

\frac{dy}{dx}

. The slope of the line segment joining

(x, y)

(-4, -3)

\frac{y + 3}{x + 4}

. Thus,

\frac{dy}{dx} = 2 \cdot \frac{y + 3}{x + 4}

. Solving this differential equation with the initial condition

(-2, 1)

, we find the equation of the curve to be

y = x^2 + 3x + 1

x^2 - 3x + 3 = 0

\sin x + \cos x = 0

\frac{dy}{dx} = x^2

x + y = 7

Correct Answer: C

Solution:

A differential equation involves derivatives of a function. Option C,

\frac{dy}{dx} = x^2

, is a differential equation because it involves the derivative of

y

with respect to

x

(x^2 + xy) dy = (x^2 + y^2) dx

y dx + x dy = 0

x dy - y dx = 0

x^2 dy + y^2 dx = 0

Correct Answer: A

Solution:

A differential equation is homogeneous if it can be expressed in the form

M(x, y) dx + N(x, y) dy = 0

where both

M(x, y)

and

N(x, y)

are homogeneous functions of the same degree. The equation

(x^2 + xy) dy = (x^2 + y^2) dx

is homogeneous because both

x^2 + xy

and

x^2 + y^2

are homogeneous functions of degree 2.

y = C

y = x + C

y = x^2 + C

y = e^x + C

Correct Answer: A

Solution:

\frac{dy}{dx} = 0

, then

y

is a constant function. Thus, the general solution is

y = C

y = C \cos(x)

y = C \sec(x)

y = C \sin(x)

y = C \cot(x)

Correct Answer: B

Solution:

The differential equation is separable. Integrating both sides, we get

\int \frac{1}{y} \, dy = \int \tan(x) \, dx

. This results in

\ln|y| = -\ln|\cos(x)| + C

, or

y = C \sec(x)

y = x - 1

y = x + 1

y = -x - 1

y = -x + 1

Correct Answer: D

Solution:

Substituting

x=0

and

y=-1

into the general solution

y = -x + C

, we find

C = -1

. Thus, the particular solution is

y = -x + 1

y = 5x^2 + 2

y = \frac{5}{2}x^2 + 2

y = \frac{5}{2}x^2

y = 5x^2

Correct Answer: B

Solution:

Integrating

\frac{dy}{dx} = 5x

gives

y = \frac{5}{2}x^2 + C

. Using the condition

y = 2

when

x = 0

, we find

C = 2

. Thus, the particular solution is

y = \frac{5}{2}x^2 + 2

y^2 = x^2 + C

y^2 = x^2 - C

y = x + C

y = x - C

Correct Answer: A

Solution:

Separating variables, we have

y dy = x dx

. Integrating both sides, we get

\frac{y^2}{2} = \frac{x^2}{2} + C

, or

y^2 = x^2 + C

. Thus, the general solution is

y^2 = x^2 + C

In the past, the solution of a differential equation was referred to as the 'integral' of the differential equation.

Correct Answer: True

Solution:

The differential equation

Solution:

A homogeneous differential equation can be solved by making the substitution

y = vx

, where

v

is a function of

x

Correct Answer: False

Solution:

Differential equations are applied in a variety of fields, including Biology, Anthropology, Geology, and Economics, not just Physics and Chemistry.

Correct Answer: False

Solution:

The degree of a differential equation is defined only if the equation is a polynomial in its derivatives, as stated in the excerpt.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

The differential equation

Solution:

Joseph Louis Lagrange first used the term 'solution' for differential equations in 1774.

Differential Equations

AI Learning Assistant

Summary

Chapter 9: Differential Equations

Summary

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 9: Differential Equations

9.1 Introduction

9.2 Basic Concepts

9.2.1 Order of a Differential Equation

9.2.2 Degree of a Differential Equation

9.3 General and Particular Solutions of a Differential Equation

9.4 Methods of Solving Differential Equations

9.5 Applications of Differential Equations

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Differential Equations

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.A spherical balloon is being inflated such that its surface area increases at a constant rate. If initially its radius is 2 units and after 4 seconds it is 4 units, what will be the radius after ttt seconds?

Correct Answer: B

Q2.What is the order of the differential equation d3ydx3+4d2ydx2+y=0\frac{d^3y}{dx^3} + 4\frac{d^2y}{dx^2} + y = 0dx3d3y​+4dx2d2y​+y=0?

Correct Answer: C

Q3.Which of the following is a homogeneous differential equation?

Correct Answer: A

Q4.Which of the following is the general solution of the differential equation dydx+ytan⁡(x)=sec⁡2(x)\frac{dy}{dx} + y \tan(x) = \sec^2(x)dxdy​+ytan(x)=sec2(x)?

Correct Answer: A

Q5.Which of the following represents a second-order differential equation?

Correct Answer: A

Q6.For the differential equation y′=exsin⁡(x)y' = e^x \sin(x)y′=exsin(x), find the equation of the curve passing through the point (0, 0).

Correct Answer: A

Q7.Which of the following represents a first-order differential equation?

Correct Answer: B

Q8.Which mathematician is credited with advocating the use of the word 'solution' for differential equations?

Correct Answer: C

Q9.Find the degree of the differential equation (d2ydx2)3+y′=0\left( \frac{d^2y}{dx^2} \right)^3 + y' = 0(dx2d2y​)3+y′=0.

Correct Answer: C

Q10.In a hypothetical chemical reaction, the rate of reaction is proportional to the square of the concentration of a reactant. If the initial concentration is 1 mol/L and it reduces to 0.5 mol/L in 2 hours, what is the concentration after 4 hours?

Correct Answer: A

Q11.The solution to the differential equation dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2 given that y=1y = 1y=1 when x=0x = 0x=0 is:

Correct Answer: A

Q12.Determine the order and degree of the differential equation (d2ydx2)3+(dydx)2=y3\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 = y^3(dx2d2y​)3+(dxdy​)2=y3.

Correct Answer: A

Q13.What is the order of the differential equation d3ydx3+2d2ydx2−5=0\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - 5 = 0dx3d3y​+2dx2d2y​−5=0?

Correct Answer: C

Q14.What is the general solution of the differential equation dy/dx=ytan⁡xdy/dx = y \tan xdy/dx=ytanx?

Correct Answer: A

Q15.What is the general solution of the differential equation dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2?

Correct Answer: A

Q16.Consider the differential equation dydx=ytan⁡(x)\frac{dy}{dx} = y \tan(x)dxdy​=ytan(x). If the solution passes through the point (0, 1), what is the particular solution?

Correct Answer: A

Q17.Which of the following is a polynomial differential equation?

Correct Answer: A

Q18.Find the general solution of the differential equation ydx−xdy=0y dx - x dy = 0ydx−xdy=0.

Correct Answer: A

Q19.Which of the following statements is true about the solution of a differential equation?

Correct Answer: B

Q20.A spherical balloon is being inflated such that its volume changes at a constant rate of 10 cubic units per second. If the initial radius is 3 units, what will be the radius after 5 seconds?

Correct Answer: B

Q21.Which of the following functions is a solution to the differential equation y′′−4y′+4y=0y'' - 4y' + 4y = 0y′′−4y′+4y=0?

Correct Answer: A

Q22.Which of the following is a particular solution to the differential equation dy/dx=ytan⁡xdy/dx = y \tan xdy/dx=ytanx given that y=1y = 1y=1 when x=0x = 0x=0?

Correct Answer: A

Q23.What is the order of the differential equation y′′′+3y′′−2y′+y=0y''' + 3y'' - 2y' + y = 0y′′′+3y′′−2y′+y=0?

Correct Answer: C

Q24.Which of the following differential equations is of degree 2?

Correct Answer: D

Q25.A curve passes through the point (0, -2) and the slope of the tangent at any point (x,y)(x, y)(x,y) on the curve is given by dydx=yx+x\frac{dy}{dx} = \frac{y}{x} + xdxdy​=xy​+x. What is the equation of the curve?

Correct Answer: A

Q26.Find the particular solution of the differential equation dydx=2x\frac{dy}{dx} = 2xdxdy​=2x given that y=3y = 3y=3 when x=1x = 1x=1.

Correct Answer: A

Q27.The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, what is the radius of the balloon after ttt seconds?

Correct Answer: C

Q28.Find the particular solution of the differential equation dydx=ycot⁡(x)\frac{dy}{dx} = y \cot(x)dxdy​=ycot(x), given that y=2y = 2y=2 when x=π4x = \frac{\pi}{4}x=4π​.

Correct Answer: A

Q29.If the solution of the differential equation dydx=x−yx+y\frac{dy}{dx} = \frac{x-y}{x+y}dxdy​=x+yx−y​ passes through the point (1, 1), what is the particular solution?