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

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)

Calculation:

Given differential equation, M dx + N dy = 0

For the differential equation to be exact, then:

⇒ My = Nx

⇒ My - Nx = 0

∴ The differential equation M dx + N dy = 0 will be exact if and only if My - Nx = 0.

The correct answer is Option 2.

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.

Calculation:

Given differential equation, M dx + N dy = 0

For the differential equation to be exact, then:

⇒ M_y = N_x

⇒ M_y - N_x = 0

∴ The differential equation M dx + N dy = 0 will be exact if and only if My - Nx = 0.

The correct answer is Option 2.

Concept:

Variable Seperable Method:

Consider the first-order differential equation, 

P(y) = Q(x), where Q(x) and P(y) are functions involving x and y only, respectively.

We can solve this by separating variables:

P(y) = Q(x) ⇒ 

Calculation:

Given, ​​​​

Let, x + y = t

⇒ 1 +  = 

⇒  =  - 1

∴ The differential equation becomes, 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Integrating on both sides, we get:

⇒ 

⇒ 

⇒ x + y - a tan^-1() = x + c

⇒ y - a tan-1() = c

⇒ tan-1() = 

⇒  = tan()

⇒ y + x = a tan(), where c is the constant of integration. 

∴ The solution of the given differential equation is (y + x) = a tan.

The correct answer is Option 1.

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:

​​​​

Integrating factor of the above differential equation is given by

I.F =

Calculation:

Given:

∴ 

P(x) = 

∴ I.F = = 

     ---(1)

The given equation can be solved using the variable separable method as:

Integrating both sides, we get:

     ---(2)

Now, from equation (1), we get:

 for y = 1

∴ y = constant = 1 is also a solution to the given differential equation.

Concept:

⇒ Q(t) e^t = e^t + C

At t = 0 Q = 0

0 = 1 + C

C = -1

Given D.E

Integrating on both sides; then

⇒ 

⇒ 

Explanation:

 (variables separable form)

Integrating both side we get,

         _______________(1)

initial condition: y = 2 at t = 0,

from equation (1), 

now at t = 3, 

Concept:

Variable separable method

∫f(y)dy = ∫f(x)dx

Calculation:

Given:

y³dy = -x³dx

Integrating both sides

At y(0) = 1

y⁴ = -x⁴ + 1

At y(-1)

y⁴ = -(-1)⁴ + 1 = 0

Concept:

Given y (0) = 1,      y(-1) = ?

y³ dy = -x³dx

      ----(1)

Now given y (0) = 1

                       {By equation (1)}

Y (-1) =?

y (-1) = 0

Integrating both sides w.r.t x

Again integrating both sides w.r.t x

y = -x⁴ + 4x³ – 10x² + c₁x + c₂                                    ……….(i)

At x = 0 and y = 5

5 = c₂

At x = 2, y =21

y = -x4 + 4x3 – 10x2 + c1x + c2   

21 = – 16 + 32 – 40 + 2c₁ + c₂

2c₁ = 21 + 16 – 32 + 40 – 5

2c₁ = 40

c₁ = 20

y = – x⁴ + 4x³ – 10 x² + 20x + 5  

Put x = 1

y = – 1 + 4 – 10 + 20 + 5

y = 18