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
Answer (Detailed Solution Below)
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
So, (ii) is exact and therefore
(2): Multiplying (i) by
So, (iii) is exact and therefore
(3): Multiplying (i) by
So, (iv) is exact and therefore
(4): Multiplying (i) by
So, (v) is not exact and therefore
(4) is correct answer.
Differential Equations Question 2:
The partial differential equation obtained by eliminating φ from :
φ(x + y + z, x2 + y2 − z2) = 0 is :
Answer (Detailed Solution Below)
Differential Equations Question 2 Detailed Solution
Concept:
The given function is:
To eliminate
Calculation:
Let
Differentiate
\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial x}) + \frac{\partial \phi}{\partial v}(2x - 2z \cdot \frac{\partial z}{\partial x}) = 0 \)
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
\( A(1 + p) + B(2x - 2z p) = 0 \ \ \ \text{(1)} \)
\( A(1 + q) + B(2y - 2z q) = 0 \ \ \ \text{(2)} \)
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 :
Answer (Detailed Solution Below)
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:
+
Answer (Detailed Solution Below)
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
C+iS = - log(
C+iS = - log(
C+iS =
Here, C=0, S=
Hence, (4) option is true.
Differential Equations Question 5:
_________ of differential equation
Answer (Detailed Solution Below)
Differential Equations Question 5 Detailed Solution
Concept:
If
Explanation:
Since,
⇒
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
Answer (Detailed Solution Below)
Differential Equations Question 6 Detailed Solution
Download Solution PDFExplanation:
3-D heat equation is given as below
For 1 – D & without heat generation:
Where α ÷ thermal diffusivity.
Wave equation is given by:
Laplace equation:
Poisson’s equation:
The differential equation 2y dx – (3y – 2x) dy = 0 is
Answer (Detailed Solution Below)
Differential Equations Question 7 Detailed Solution
Download Solution PDFConcept:
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
As
so, it is an exact equation.
If y = e3x + e-5x find the value of
Answer (Detailed Solution Below)
Differential Equations Question 8 Detailed Solution
Download Solution PDFConcept:
Calculation:
Given:
y = e3x + e-5x
At x = 0
we have
For the equation
Answer (Detailed Solution Below)
Differential Equations Question 9 Detailed Solution
Download Solution PDFConcept:
For solving first order, first-degree differential equations always first inspect with variable separation method.
Calculation:
Given the differential equation is,
Where A is a constant.
Use condition
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
Answer (Detailed Solution Below)
Differential Equations Question 10 Detailed Solution
Download Solution PDFConcept:
Integrating factor of the above differential equation is given by
I.F =
Calculation:
Given:
∴
P(x) =
∴ I.F =
The partial differential equation
Answer (Detailed Solution Below)
Differential Equations Question 11 Detailed Solution
Download Solution PDFExplanation:
Order of a differential equation
Order is defined by the highest derivative present in the equation.
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
The integrating factor of the differential equation
Answer (Detailed Solution Below)
Differential Equations Question 12 Detailed Solution
Download Solution PDFConcept:
Equation of this type is known as Linear First Order Equation, whose solution is given by:
Calculation:
Given:
By comparing equation (1) with
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
Answer (Detailed Solution Below)
Differential Equations Question 13 Detailed Solution
Download Solution PDFConcept:
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
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Differential Equations Question 15 Detailed Solution
Download Solution PDFExplanation:
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;
∴