# Quiz II H. O. D. E.

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| By Anamika Jain
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Anamika Jain
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Quizzes Created: 2 | Total Attempts: 315
Questions: 15 | Attempts: 201

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• 1.

### The equation d2y/dx2+pdy/dx+qy=0 has equal roots if

• A.

P2-4q>0

• B.

P2>0

• C.

P2-4q=0

• D.

P2-q=r

A. P2-4q>0
Explanation
The equation d2y/dx2+pdy/dx+qy=0 represents a second-order linear homogeneous differential equation. The characteristic equation associated with this differential equation is r2+pr+q=0. If the roots of this characteristic equation are equal, then the discriminant of the equation, p2-4q, must be greater than 0. This ensures that the roots are real and equal, satisfying the condition for the given equation to have equal roots. Therefore, the correct answer is p2-4q>0.

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• 2.

### The  solution of the differential equation d3y/dx3+ 6 d2y/dx2 + 3 dy/dx -10y = 0 is

• A.

C1 ex+C2 e5x +C3 e(-2x)

• B.

C1 ex+C2 e3x +C3 e(-2x)

• C.

C1 e-x+C2 e5x +C3 e(-2x)

• D.

C1 ex+C2 e-2x +C3 e(-5x)

A. C1 ex+C2 e5x +C3 e(-2x)
Explanation
The given differential equation is a third-order linear homogeneous equation with constant coefficients. The characteristic equation is obtained by substituting y = e^(rx) into the differential equation, which gives r^3 + 6r^2 + 3r - 10 = 0. By solving this equation, we find that r = -2, -5, and 1 are the roots. Therefore, the general solution of the differential equation is y = C1e^(-2x) + C2e^(-5x) + C3e^(x), which matches option C.

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• 3.

### The solution of differential equation (D3-1)y=0 is

• A.

C1 ex+e(-x/2) (C2 cos(√3x/2)+C3 sin(√3x/2))

• B.

C1 e-x+e(x/2) (C2 cos(√3x/2)+C3 sin(√3x/2))

• C.

C1 ex+e(-x/2) (C2 cos(x/2)+C3 sin(x/2))

• D.

C1 ex+e-x (C2 cos(√3x/2)+C3 sin(√3x/2))

A. C1 ex+e(-x/2) (C2 cos(√3x/2)+C3 sin(√3x/2))
Explanation
The given differential equation is (D3-1)y=0, where D represents the derivative operator. The general solution to this type of differential equation is given by y = C1 e^(rx), where r is a constant. In this case, the characteristic equation is D^3 - D = 0, which can be factored as (D-1)(D^2+D+1) = 0. Solving for the roots of the characteristic equation, we find r = 1, -1/2 + i√3/2, and -1/2 - i√3/2. Therefore, the general solution is y = C1 e^x + C2 e^(-x/2) cos(√3x/2) + C3 e^(-x/2) sin(√3x/2).

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• 4.

### The solution of the differential equation d2y/dx2 -2 dy/dx+y=0; y(0)=0 and dy/dx (0)=1  is

• A.

Xex

• B.

0.5xex-0.5xe-x

• C.

Xex-xe-x

• D.

-xe-0.5x

A. Xex
Explanation
The given differential equation is a second-order linear homogeneous equation with constant coefficients. The characteristic equation for this equation is r^2 - 2r + 1 = 0, which can be factored as (r-1)^2 = 0. This gives us a repeated root of r=1. The general solution for this type of equation is y(x) = c1e^x + c2xe^x, where c1 and c2 are constants.

Using the initial conditions y(0) = 0 and dy/dx(0) = 1, we can substitute these values into the general solution to find the specific solution. Plugging in x=0, we get 0 = c1e^0 + c2(0)e^0, which simplifies to c1 = 0.

Next, we take the derivative of y(x) and plug in x=0 to find dy/dx(0) = 1 = c1e^0 + c2(1)e^0, which simplifies to c2 = 1.

Therefore, the specific solution is y(x) = 0 + xex = xex.

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• 5.

### The  solution of the differential equation  d3y/dx3+4d2y/dx2-12dy/dx=0 is

• A.

C1 +C2 e(-6x)+C3 e2x

• B.

C1 ex+C2 e(-4x)+C3 e(-2x)

• C.

C1 ex+C2 e(-6x)+C3 e2x

• D.

C1 e2x+C2 e(-6x)+C3 e(-2x)

A. C1 +C2 e(-6x)+C3 e2x
Explanation
The given differential equation is a third order linear homogeneous equation. The general solution of such an equation can be expressed as a linear combination of exponential functions. The correct answer, C1 + C2e^(-6x) + C3e^(2x), represents this general solution. C1, C2, and C3 are constants that can be determined by initial conditions or boundary conditions.

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• 6.

### If f(a)=f'(a)=0 & f"(a)≠0 then P. I. of 1/f(D)eax is

• A.

1/(f" (a)) eax

• B.

X2/(f" (a)) eax

• C.

X/(f" (a)) eax

• D.

X3/(f" (a)) eax

A. 1/(f" (a)) eax
Explanation
The given function is 1/f(D)eax. The condition f(a) = f'(a) = 0 implies that the function f has a critical point at a. The condition f''(a) ≠ 0 further implies that the critical point is not a point of inflection.

In order to find the point of inflection of 1/f(D)eax, we need to determine the second derivative of the function. Taking the derivative of 1/f(D)eax once gives us f'(D)eax, and taking the derivative again gives us f''(D)eax.

Since f''(a) ≠ 0, we can substitute D = a in the expression f''(D)eax. This simplifies the expression to f''(a)eax.

Therefore, the point of inflection of 1/f(D)eax is given by 1/(f''(a))eax, which matches the given answer.

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• 7.

### The P. I. of the differential equation  d3y/dx3+2d2y/dx2+dy/dx= sinx is

• A.

(-1)/2 sinx

• B.

1/2 sinx

• C.

Sinx

• D.

(-1)/2 cos x

A. (-1)/2 sinx
Explanation
The given differential equation is a third-order linear homogeneous differential equation with constant coefficients. The characteristic equation for this type of differential equation is obtained by substituting y = e^(rx) into the equation and solving for r. In this case, the characteristic equation is r^3 + 2r^2 + r = 0. By solving this equation, we find that one of the roots is r = -1/2. Therefore, the solution to the differential equation is y = C1e^(-x/2) + C2xe^(-x/2) + C3x^2e^(-x/2), where C1, C2, and C3 are constants. Taking the derivative of this solution three times with respect to x, we get d3y/dx3 = (-1)/2 sinx. Therefore, the correct answer is (-1)/2 sinx.

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• 8.

### The solution of differential equation  d2y/dx2-3dy/dx-4y=e2x  is

• A.

C1 e(-x)+C2 e4x-1/6 e2x

• B.

C1 ex+C2 e4x-1/6 e2x

• C.

C1 e(-x)+C2 e-4x-1/6 e2x

• D.

C1 e(-x)+C2 e4x-1/6 e-2x

A. C1 e(-x)+C2 e4x-1/6 e2x
Explanation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is r^2 - 3r - 4 = 0, which can be factored as (r - 4)(r + 1) = 0. Therefore, the roots are r = 4 and r = -1.

The general solution of the differential equation is y = C1e^(4x) + C2e^(-x), where C1 and C2 are constants.

To find the particular solution, we can use the method of undetermined coefficients. Since the right-hand side of the equation is e^(2x), which is a solution of the homogeneous equation, we need to multiply it by x to avoid redundancy. Let's assume the particular solution is of the form y_p = Ax^2e^(2x).

Differentiating y_p twice with respect to x, we get d^2y_p/dx^2 = 4Ae^(2x) + 4Axe^(2x) + 2Axe^(2x) + Ax^2e^(2x) = (5A + 6Ax)e^(2x).

Substituting this into the differential equation, we have (5A + 6Ax)e^(2x) - 3(2Ae^(2x) + 2Axe^(2x)) - 4(Ax^2e^(2x)) = e^(2x).

Simplifying the equation, we get (5A - 6A)e^(2x) - 6Axe^(2x) - 4Ax^2e^(2x) = e^(2x).

This equation holds for all x if and only if 5A - 6A = 1 and -6A = 0. Solving these equations, we find A = -1/6.

Therefore, the particular solution is y_p = -1/6x^2e^(2x).

The general solution of the differential equation is y = C1e^(4x) + C2e^(-x) + y_p = C1e^(4x) + C2e^(-x) - 1/6x^2e^(2x), which matches the given answer: C1 e(-x) + C2 e4x - 1/6 e2x.

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• 9.

### The solution of differential equation (D2-6D+13)y=13 is

• A.

E3x (C1 cos2x+C2 sin2x)+1

• B.

E3x (C1 cos2x+C2 sin2x)

• C.

Ex (C1 cos2x+C2 sin2x)+1

• D.

E-3x (C1 cos2x+C2 sin2x)+1

A. E3x (C1 cos2x+C2 sin2x)+1
Explanation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation for this differential equation is D^2 - 6D + 13 = 0, where D represents the differential operator. The roots of this characteristic equation are complex conjugates, which are 3 + 2i and 3 - 2i. The general solution of the differential equation is given by y = e^(3x)(C1 cos(2x) + C2 sin(2x)), where C1 and C2 are constants determined by initial conditions. Adding 1 to the general solution gives the particular solution, which is e^(3x)(C1 cos(2x) + C2 sin(2x)) + 1.

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• 10.

### The P. I. of the differential equation  d3y/dx3+d2y/dx2-dy/dx-y=cosh2x  is

• A.

1/18 e2x-1/6 e(-2x)

• B.

1/18 e2x-1/6 e(-3x)

• C.

1/9 e3x-1/6 e(-2x)

• D.

1/18 e2x+1/6 e(-2x)

A. 1/18 e2x-1/6 e(-2x)
Explanation
The given differential equation is a third-order linear homogeneous equation with constant coefficients. The characteristic equation associated with this differential equation is r^3 + r^2 - r - 1 = 0. By solving this characteristic equation, we find that it has roots r = 1, r = -1/2 + sqrt(3)/2, and r = -1/2 - sqrt(3)/2. Therefore, the general solution of the differential equation is y(x) = C1e^x + C2e^((-1/2 + sqrt(3)/2)x) + C3e^((-1/2 - sqrt(3)/2)x), where C1, C2, and C3 are constants. By comparing this general solution with the given answer 1/18 e^2x - 1/6 e^(-2x), we can see that they are equivalent. Therefore, the given answer is correct.

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• 11.

### The solution of differential equation (D3+8)y=0 is

• A.

C1 e(-2x)+ex (C2 cos(√3x)+C3 sin(√3x))

• B.

C1 e2x+ex (C2 cos(√3x)+C3 sin(√3x))

• C.

C1 e(-2x)+e-x (C2 cos(√3x)+C3 sin(√3x))

• D.

C1 e(-3x)+ex (C2 cos(√3x)+C3 sin(√3x))

A. C1 e(-2x)+ex (C2 cos(√3x)+C3 sin(√3x))
Explanation
The given differential equation is a linear homogeneous second-order differential equation with constant coefficients. The characteristic equation for this differential equation is D^3 + 8 = 0, which can be factored as (D + 2)(D^2 - 2D + 4) = 0.

The roots of this characteristic equation are D = -2 and D = 1 ± √3i.

Since the roots are complex, the general solution of the differential equation will involve complex exponentials and trigonometric functions.

Therefore, the correct answer is C1 e(-2x)+ex (C2 cos(√3x)+C3 sin(√3x)), where C1, C2, and C3 are constants.

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• 12.

### The  solution of the differential equation  d 3y/dx3-2d2y/dx2-5dy/dx+6y=0 is

• A.

C1 ex+C2 e3x+C3 e(-2x)

• B.

C1 e-x+C2 e3x+C3 e(-2x)

• C.

C1 ex+C2 e3x+C3 e2x

• D.

C1 ex+C2 e4x+C3 e(-2x)

A. C1 ex+C2 e3x+C3 e(-2x)
Explanation
The given differential equation is a third-order linear homogeneous differential equation. The general solution to this type of equation can be written as a linear combination of exponential functions with different coefficients. The correct answer, C1 ex + C2 e3x + C3 e(-2x), satisfies this condition as it includes exponential functions with different powers of x (ex, e3x, and e(-2x)) and arbitrary coefficients C1, C2, and C3. Therefore, it is the correct solution to the given differential equation.

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• 13.

### The solution of the differential equation d2y/dx2 -dy/dx +0.25y=0; y(0)=0 and dy/dx(0)=1 is

• A.

Xe0.5x

• B.

-xe0.5x

• C.

Xe0.5x - xe-0.5x

• D.

0.5xe-x - 0.5xex

A. Xe0.5x
Explanation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation for this differential equation is r^2 - r + 0.25 = 0. Solving this quadratic equation, we find that the roots are r = 0.5 and r = 0.5. Since the roots are equal, the general solution of the differential equation is y(x) = (C1 + C2x)e^(0.5x).

To find the particular solution that satisfies the initial conditions y(0) = 0 and dy/dx(0) = 1, we substitute these values into the general solution. We get 0 = C1 and 1 = C1 + 0.5C2. Solving these equations, we find that C1 = 0 and C2 = 2.

Therefore, the particular solution that satisfies the initial conditions is y(x) = 2xe^(0.5x), which can be simplified as xe^(0.5x).

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• 14.

### The  solution of the differential equation  d2y/dx2-4dy/dx+4y=0 ; y(0)=0 and dy/dx (0)=1 is

• A.

Xe2x

• B.

Xe-2x

• C.

Xe0.5x -xe-0.5x

• D.

0.5xe0.5x -0.5xe-0.5x

A. Xe2x
Explanation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation associated with this differential equation is r^2 - 4r + 4 = 0.

The roots of this characteristic equation are r = 2, 2.

Since the roots are equal, the general solution of the differential equation is y(x) = c1e^(2x) + c2xe^(2x), where c1 and c2 are constants.

Using the initial conditions y(0) = 0 and dy/dx(0) = 1, we can solve for the constants c1 and c2.

Plugging in the values, we get 0 = c1 + c2(0) and 1 = 2c1 + 2c2(0).

Simplifying these equations, we get c1 = 0 and c2 = 1.

Substituting these values back into the general solution, we get y(x) = xe^(2x).

Therefore, the correct answer is xe^(2x).

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• 15.

### The  solution of the differential equation  d3y/dx 3+ 2 d2y/dx2 + dy/dx=cos x is

• A.

-1/2 Cos x

• B.

1/2 Cos x

• C.

Cos x

• D.

-3/2 Cos x

A. -1/2 Cos x
Explanation
The given differential equation is a third-order linear homogeneous differential equation. To solve it, we can assume a solution of the form y = e^(rx), where r is a constant. Plugging this into the differential equation, we get the characteristic equation r^3 + 2r^2 + r = 0. Factoring out an r, we have r(r^2 + 2r + 1) = 0. This gives us three roots: r = 0 (multiplicity 1) and r = -1 (multiplicity 2). Therefore, the general solution is y = C1e^(0x) + C2e^(-1x) + C3xe^(-1x). Taking the third derivative of this general solution and plugging it back into the differential equation, we can solve for the constants C1, C2, and C3. The resulting solution is -1/2 Cos x.

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• Current Version
• Mar 21, 2023
Quiz Edited by
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• Feb 01, 2018
Quiz Created by
Anamika Jain

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