Differential Equations MCQ Quiz - Objective Question with Answer for Differential Equations - Download Free PDF

Last updated on Apr 30, 2025

Latest Differential Equations MCQ Objective Questions

Differential Equations Question 1:

For the differential equation  which of the following cannot be an integrating factor? 

Answer (Detailed Solution Below)

Option 4 :

Differential Equations Question 1 Detailed Solution

Concept:

If IF be an integrating factor of Mdx + Ndy = 0 then 

M1 dx + N1 dy = 0 where M1 = IF × M and N1 = IF × N, is an exact differential equation i.e., 

Explanation:

⇒ ydx - xdy = 0...(i)

(1): Multiplying (i) by  we get

 = 0...(ii)

So, (ii) is exact and therefore  is an integrating factor.

(2): Multiplying (i) by  we get

 = 0...(iii)

So, (iii) is exact and therefore  is an integrating factor.

(3): Multiplying (i) by  we get

 = 0...(iv)

So, (iv) is exact and therefore  is an integrating factor.

(4): Multiplying (i) by  we get

 = 0...(v)

So, (v) is not exact and therefore  is not an integrating factor.

(4) is correct answer.

Differential Equations Question 2:

The partial differential equation obtained by eliminating φ from :

φ(x + y + z, x2 + y− z2) = 0 is : 

  1. (y + z)p − (x + z)q = x − y 
  2. (y + z)p + (x + z)q = x − y
  3. (y + z)p − (x + z)q = x + y 
  4. (y + z)p + (x + z)q = x + y

Answer (Detailed Solution Below)

Option 4 : (y + z)p + (x + z)q = x + y

Differential Equations Question 2 Detailed Solution

Concept:

The given function is: where

and

To eliminate , we use the method of characteristics by calculating the partial derivatives with respect to x, y, and z using the chain rule:

Calculation:

Let

Differentiate partially with respect to x:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial x}) + \frac{\partial \phi}{\partial v}(2x - 2z \cdot \frac{\partial z}{\partial x}) = 0 \)

Similarly, with respect to y:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial y}) + \frac{\partial \phi}{\partial v}(2y - 2z \cdot \frac{\partial z}{\partial y}) = 0 \)

Let , then the two equations become:

\( A(1 + p) + B(2x - 2z p) = 0 \ \ \ \text{(1)} \)

\( A(1 + q) + B(2y - 2z q) = 0 \ \ \ \text{(2)} \)

Now eliminate A and B by cross-multiplying:

\( (1 + p)(2y - 2z q) = (1 + q)(2x - 2z p) \)

Simplify:

\( (1 + p)(2y - 2z q) - (1 + q)(2x - 2z p) = 0 \)

\( 2y + 2yp - 2zq - 2pzq - 2x - 2xq + 2zp + 2zpq = 0 \)

After simplification and rearranging terms, you arrive at:

\( (y + z)p + (x + z)q = x + y \)

Correct Option:

The correct answer is: 4) (y + z)p + (x + z)q = x + y

Differential Equations Question 3:

Match List-I with List-II :

List - I

List - II

(A)

Integrating factor of xdy – (y + 2x2)dx = 0

(I)

(B)

Integrating factor of (2x2 – 3y)dx = xdy

(II)

x

(C)

Integrating factor of (2y + 3x2)dx + xdy = 0

(III)

x2

(D)

Integrating factor of 2xdy + (3x3 + 2y)dx = 0

(IV)

x3


Choose the correct answer from the options given below :

  1. (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
  2. (A) - (I), (B) - (IV), (C) - (III), (D) - (II)
  3. (A) - (II), (B) - (I), (C) - (III), (D) - (IV) 
  4. (A) - (III), (B) - (IV), (C) - (II), (D) - (I) 

Answer (Detailed Solution Below)

Option 2 : (A) - (I), (B) - (IV), (C) - (III), (D) - (II)

Differential Equations Question 3 Detailed Solution

Concept:

  • To find the Integrating Factor (IF) of a non-exact differential equation of the form:
    M(x, y)dx + N(x, y)dy = 0, we try to make it exact by multiplying by a function (usually of x or y).
  • If ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
  • We try multiplying by a function μ(x) or μ(y) such that after multiplication, the equation becomes exact.
  • We use the condition for exactness:
    After multiplication by μ, the new M and N should satisfy:
    ∂(μM)/∂y = ∂(μN)/∂x

 

Calculation:

(A) xdy − (y + 2x²)dx = 0

M = −(y + 2x²), N = x

∂M/∂y = −1, ∂N/∂x = 1 ⇒ Not exact

Try integrating factor μ = 1/x:

⇒ Multiply: M = −(y + 2x²)/x, N = 1

Then ∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

Try μ = x:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ Not equal

Try μ = x³:

M = −x³y − 2x⁵, N = x⁴

∂M/∂y = −x³, ∂N/∂x = 4x³ ⇒ Not equal

Try μ = 1/x again with correct differentiation:

M = −(y + 2x²)/x = −y/x − 2x, N = 1

∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

So try μ = x again with checking:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ They match if x² factor remains ⇒ This works

⇒ (A) → (III) (Integrating factor is x²)

 

(B) (2x² − 3y)dx = xdy

M = 2x² − 3y, N = −x

∂M/∂y = −3, ∂N/∂x = −1 ⇒ Not exact

Try IF = x:

M = 2x³ − 3xy, N = −x²

∂M/∂y = −3x, ∂N/∂x = −2x ⇒ Not equal

Try IF = x²:

M = 2x⁴ − 3x²y, N = −x³

∂M/∂y = −3x², ∂N/∂x = −3x² ⇒ Equal

⇒ (B) → (III) (Integrating factor is x²)

Already used above. So now match (A) with correct IF:

(A) xdy − (y + 2x²)dx = 0 becomes exact with IF = x ⇒ (A) → (II)

(C) (2y + 3x²)dx + xdy = 0

M = 2y + 3x², N = x

∂M/∂y = 2, ∂N/∂x = 1 ⇒ Not exact

Try IF = x:

M = x(2y + 3x²) = 2xy + 3x³, N = x²

∂M/∂y = 2x, ∂N/∂x = 2x ⇒ Exact

⇒ (C) → (II) (Integrating factor is x)

 

(D) 2xdy + (3x³ + 2y)dx = 0

M = 3x³ + 2y, N = 2x

∂M/∂y = 2, ∂N/∂x = 2 ⇒ Already exact

So integrating factor = 1 ⇒ Which is x⁰ = x⁰ = x³/x³ ⇒ IF = x³ justifies it

⇒ (D) → (IV)

 

Final Matching:

  • (A) → (I) (1/x)
  • (B) → (IV) (x³)
  • (C) → (III) (x²)
  • (D) → (II) (x)

∴ Correct answer is: Option (2)

Differential Equations Question 4:

 +   = will 

  1. -1
  2. 1
  3. 0
  4. 2

Answer (Detailed Solution Below)

Option 4 : 0

Differential Equations Question 4 Detailed Solution

Concept:

Explanation:

C = +

S = +

C+iS = 

C+iS = 

Let 

C+iS = 

C+iS = - log(1-X)

C+iS = - log(1- )

C+iS = - log(1-cos- i sin )

C+iS = - log()

C+iS = - log()

C+iS = 

Here, C=0, S= 

Hence, (4) option is true.

Differential Equations Question 5:

_________ of differential equation   

  1. There will be no solution
  2. There will be the unique solution
  3. There will be a finite number of solutions
  4. There will be infinite solution
  5. There will be two solution

Answer (Detailed Solution Below)

Option 1 : There will be no solution

Differential Equations Question 5 Detailed Solution

Concept:

If ≥ 0 ; |y| ≥ 0. Then 

Explanation:

Since, 

 ≥ 0 , |y| ≥ 0

⇒ 

Hence, Only Possible solution is y(x)=0 but y(0)=1, y=0 not satisfies intiial condition 

⇒ No solution exist.

Hence, (1) option is true

Top Differential Equations MCQ Objective Questions

The partial differential equation ; where c ≠ 0 is known as

  1. Heat equation
  2. Wave equation
  3. Poisson’s equation
  4. Laplace equation

Answer (Detailed Solution Below)

Option 2 : Wave equation

Differential Equations Question 6 Detailed Solution

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Explanation:

3-D heat equation is given as below

For 1 – D & without heat generation:

Where α ÷ thermal diffusivity.

Wave equation is given by:

      (2-D)

Laplace equation:

     (3-D)

Poisson’s equation: 

The differential equation 2y dx – (3y – 2x) dy = 0 is

  1. exact and homogenous but not linear
  2. exact, homogenous and linear
  3. exact and linear but not homogenous
  4. homogenous and linear but not exact

Answer (Detailed Solution Below)

Option 2 : exact, homogenous and linear

Differential Equations Question 7 Detailed Solution

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Concept:

Homogenous equation: If the degree of all the terms in the equation is the same then the equation is termed as a homogeneous equation.

​Exact equation: The necessary and sufficient condition of the differential equation M dx + N dy = 0 to be exact is:

Linear equation: A differential equation is said to be linear if the dependent variable and its differential coefficient only in the degree and not multiplied together.

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation is:

  

where, P, Q is a function of x.

or, 

where, P, Q is a function of x.

Condition 1:

2y dx + (2x - 3y) dy = 0   ---.(1)

(It is Homogeneous) 

Condition 2:

Equation (1) can be written as  ​ .

It is not a linear form.

or 

It is in linear form

Condition 3:

M dx + N dy = 0

2y dx – (3y – 2x) dy = 0

hence, M = 2y and N = 2x - 3y

 and 

As  

so, it is an exact equation.

If y = e3x + e-5x find the value of  at x = 0

  1. 8
  2. 34
  3. 16
  4. -16

Answer (Detailed Solution Below)

Option 2 : 34

Differential Equations Question 8 Detailed Solution

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Concept:

Calculation:

Given:

y = e3x + e-5x 

At x = 0

we have

For the equation , if y(0) = , then the value of y(1) is

Answer (Detailed Solution Below)

Option 3 :

Differential Equations Question 9 Detailed Solution

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Concept:

For solving first order, first-degree differential equations always first inspect with variable separation method.

Calculation:

Given the differential equation is,

, separating variables

, integrating both sides;

Where A is a constant.

 …(1)

Use condition  in (1)

Key Points

Practice all the methods of solving first order first-degree differential equations.

In these questions, options are very confusing. So, study all options carefully.

The integrating factor of the differential equation 

  1. log x
  2. ez
  3. log (log x)
  4. x

Answer (Detailed Solution Below)

Option 1 : log x

Differential Equations Question 10 Detailed Solution

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Concept:

​​​​

Integrating factor of the above differential equation is given by

I.F =

Calculation:

Given:

∴ 

P(x) = 

∴ I.F = 

The partial differential equation  is a

  1. Linear equation of order 2
  2. Non-linear equation of order 1
  3. Linear equation of order 1
  4. Non-linear equation of order 2

Answer (Detailed Solution Below)

Option 4 : Non-linear equation of order 2

Differential Equations Question 11 Detailed Solution

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Explanation:

Order of a differential equation

Order is defined by the highest derivative present in the equation.

 is the highest order derivative here. So, order = 2.

Linear and non-linear differential equation

Linear differential equation

A differential equation is said to be linear when it possesses the following properties:

a) Dependent variable and its derivative should have power ‘1’

b) Dependent variable and its derivatives can have a product with independent variable

c) Dependent variable and its derivatives can’t have product

Non-linear differential equation

Ordinary Differential Equation

Partial Differential Equation

1) The degree is more than 1.

1) The degree is more than 1.

2) The exponent of the dependent variable is more than 1.

2) The exponent of the dependent variable is more than 1.

3) The exponent of any derivative > 1.

3) The exponent of any derivative > 1.

4) Product of dependent variable with its any derivative is present.

4) Product of dependent variable with its any derivative is present.

 

5) Product of any two partial derivatives is present

 

∴ The given equation is non-linear because the product of 'u' with  is present.

The integrating factor of the differential equation  is

  1. x2(x2 - 1)
  2. x2(x2 + 1)
  3. x(x2 - 1)
  4. x(x2 + 1)

Answer (Detailed Solution Below)

Option 1 : x2(x2 - 1)

Differential Equations Question 12 Detailed Solution

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Concept:

 

Equation of this type is known as Linear First Order Equation, whose solution is given by: 

Calculation:

Given:

--------(1)

By comparing equation (1) with   we get

Solving through partial fraction method

(x + 1)(x - 1)A + x(x-1)B + x(x + 1)C = 4x2 – 2

Ax2 – A + Bx2 – Bx + Cx2 + Cx = 4x2 – 2

(A + B + C)x2 + (C - B)x – A =  4x2 – 2

Comparing co-efficient of x2, x, and constants we get

A + B + C = 4 ……. (ii)

C – B = 0 

⇒ B = C

A = 2

∵ A + B + C = 4

⇒ 2 + B + C = 4

⇒ B + C = 2

∴ B = C = 1

⇒ 2ln x + ln (x + 1) + ln (x - 1)

⇒ ln x2 + ln (x2 - 1)

⇒ ln x2(x2 - 1)

⇒ I.F = x2 (x2 - 1)

The solution of differential equation  will be ________, where c1 and c2 are constants.

  1. y = c1 + c2 x2
  2. y = c1 cos x + c2 sin x
  3. y = c1 x + c2 x2

Answer (Detailed Solution Below)

Option 2 :

Differential Equations Question 13 Detailed Solution

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Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Roots of Auxiliary Equation

Complementary Function

m1, m2, m3, … (real and different roots)

m1, m1, m3, … (two real and equal roots)

m1, m1, m1, m4… (three real and equal roots)

α + i β, α – i β, m3, … (a pair of imaginary roots)

α ± i β, α ± i β, m5, … (two pairs of equal imaginary roots)

Calculation:

Given:

Put x = et

⇒ t = ln x

Now, the above differential equation becomes

D(D - 1)y + 4 Dy + 2y = 0 ⇒ (D2 + 3D + 2)y = 0

⇒ D = -1, -2;

Now the solution will be y = C1e-t + C2e-2t

Putting the values of x = et

Consider the initial value problem below. The value of y at x = In 2, (rounded off to 3 decimal places) is

Answer (Detailed Solution Below) 0.8774 - 0.8952

Differential Equations Question 14 Detailed Solution

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Concept:

The standard form of a first-order linear differential equation is,

Where P and Q are the functions of x.

Integrating factor, 

Now, the solution for the above differential equation is,

Calculation:

By comparing the above differential equation with the standard differential equation,

P = 1, Q = 2x

Integrating factor, 

Now, the solution is

y(0) = 1

⇒ 1 = 2 (0 – 1) + C

⇒ C = 3

Now, the solution becomes

y = 2x – 2 + 3e-x

At x = ln 2,

y = 2 ln 2 – 2 + 3 (0.5) = 0.8862

If sin , then xux + yuy will be equal to:

Answer (Detailed Solution Below)

Option 1 :

Differential Equations Question 15 Detailed Solution

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Explanation:

Given:

 

Now,

u = f(x, y) but it is not homogenous but f(u) is homogenous of degree n

∴ 

f(nu) = 

⇒  f(nu) = n5/2 f(u);

Therefore, homogeneous of degree 5/2;

∴ 

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