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

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.

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

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)

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.

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

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: 

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.

Concept:

Calculation:

Given:

y = e3x + e-5x 

At x = 0

we have

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.

Concept:

​​​​

Integrating factor of the above differential equation is given by

I.F =

Calculation:

Given:

∴ 

P(x) = 

∴ I.F = = 

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

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

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)

Concept:

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

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 = e^t

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

Explanation:

Given:

Now,

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

∴ 

f(nu) = 

⇒  f(nu) = n^5/2 f(u);

Therefore, homogeneous of degree 5/2;

∴