Solving Linear Differential Equation MCQ Quiz - Objective Question with Answer for Solving Linear Differential Equation - Download Free PDF

If 

 Let 

Let  

 .........(I)

At 

At 

Explanation:

A.  

This function is defined for x > 0, as the logarithm function is only defined for positive values of x.

To find the interval in which the function is increasing, take the derivative of    with respect to x.

After solving, it can be found that the function is increasing when x > e.

So, the correct interval is (IV).

B.   

This function is defined for  

By finding the derivative and setting it to zero,

it can be shown that the function is increasing for x > 2 and x

So, the correct interval is (I).

C.   0 \)

This function is increasing for x > 0.

So, the correct interval is (III).

D. 

The derivative of    is   

Setting this derivative greater than zero gives the interval   , so the correct interval is (II).

Final Answer:

The correct matching is:

(A) → (IV)
(B) → (I)
(C) → (III)
(D) → (II)

The correct option is (3).

Concept:

Integrating Factor:

• The integrating factor is used to solve linear first-order differential equations.
• For a linear differential equation of the form: dy/dx + P(x)y = Q(x), the integrating factor is given by:
• Integrating Factor = e^∫P(x)dx

Calculation:

Given the differential equation:

y log_e(y) dx/dy + x = 2 log_e(y)

Rewriting the equation:

y log_e(y) dx/dy = 2 log_e(y) - x

Dividing both sides by y log_e(y):

dx/dy = 2/y - x/(y log_e(y))

This is in the standard form of a linear first-order differential equation:

dx/dy + P(y) x = Q(y)

Identifying P(y) = 1/(y log_e(y)) and Q(y) = 2/y, we find the integrating factor:

μ(y) = e^∫P(y)dy = e^{∫1/(y log_e(y))dy}

The integral of 1/(y log_e(y)) is log_e(log_e(y)), so:

μ(y) = log_e(y)

Hence, the integrating factor is: log_e(y)

Concept:

Integration and Constant of Integration:

• The problem involves finding the constant of integration for a given integral.
• The integral is of the form:
• To solve this, we use the substitution method to simplify the integral.
• The substitution used is:
  • , and hence
• The integral then reduces to a standard form:
  • ∫ 1 / √(4 - u²) du =   
• We substitute back to get the final result:
  •   
• To find the constant C, we use the initial condition that f(0) = π/2.

Calculation:

Given,

Substitute  , so that   .

The integral becomes  

Use the initial condition:

f(0) = π/2

Substituting x = 0 into the equation:

Set this equal to π/2:

π/6 + C = π/2

Solve for C:

C = π/2 - π/6 = 3π/6 - π/6 = 2π/6 = π/3

Hence, the constant of integration is: C = π/3

Given:

x = 1

x2 + y2 + z2 = xy + yz + zx

Calculations:

x² + y2 + z2 - xy - yz - zx = 0

⇒(1/2)[(x - y)² + (y - z)² + (z - x)²] = 0

⇒x = y , y = z and z = x

But x = y = z = 1

so, 

= {10(1)⁴ + 5(1)⁴ + 7(1)⁴}/{13(1)²(1)²+ 6(1)²(1)² + 3(1)²(1)²}

= 22/22

= 1

Hence, the required value is 1.

Given:

x +  = 3

Concept Used:

Simple calculations is used

Calculations:

⇒ x +  = 3

On multiplying 2 on both sides, we get

⇒ 2x +  = 6  .................(1)

Now, On cubing both sides,

⇒ 

⇒ 

⇒ 

⇒ 

⇒   ..............from (1)

⇒ 

⇒ 

⇒ Hence, The value of the above equation is 180

Given:

9-digit number = 83P93678Q

Divisor = 72

Concept Used:

Divisibility of 8 = Last three digits should be divisible by 8.

Divisibility of 9 = Sum of digits is divisible by 9.

Calculation:

As the divisor 72, is divisible by 8 and 9, so the divisibility will be checked.

For divisible by 8,

78Q should be divisible by 8, so, Q should be 4 as 784 is divisible by 8.

For divisible by 9,

⇒ 8 + 3 + P + 9 + 3 + 6 + 7 + 8 + 4 = 48 + P

For being divisible by 9, the nearest number to be added is 6 which gives 54.

Now, 

⇒ 

Therefore, the required value is 8.

Concept: 

Integrating factor, (IF) for a differential equation, , where P and  Q are given continuous function of y. 

IF =  

Calculation:

Given differential equation

Now, this differential equation is in the form

where, P(x) = x and Q(x) = x

Integrating Factor (I.F.) = 

I.F. =  =  

Concept:

First Order Linear Differential Equation:
A differential equation of the from  + P × y = Q, where P and Q are constants or functions of x only, is known as a first order linear differential equation.

Steps to solve a First Order Linear Differential Equation:

• Convert into the standard form  + P × y = Q, where P and Q are constants or functions of x only.
• Find the Integrating Factor (F) by using the formula: F = .
• Write the solution using the formula:  where C is the constant of integration.

Calculation:

⇒ P =  and Q = .

Integrating factor F = .

The solution of the given differential equation is:

⇒ y = Cx - 3, which is the equation of a straight line.

Concept:

Solution of Linear Differential equation:

If the D.E. has a form of  then, where P and Q are functions of y.

The solution is given as, 

where, I.F. is integrating factor which is given as,

Calculation:

Given: ydx – (x + 2y²) dy = 0

Differential equation is in form of, 

Integrating factor, 

Differential equation is given as,

⇒ x = 2y² + cy

Concept:

The solution of the linear differential equation  is given by

 y × I.F = 

Where P and Q are the functions of 'x' and I.F =  

Calculation:

Given  ⇒   + y -  = 

⇒   + y = 

This is a differential equation of the form 

Here P(x) = 1 - 

Integrating factor (I.F) =   =  = 

⇒ I.F =  = 

The integral factor is .

The correct answer is option 2.

Concept:

Equation of the form  solve by following the steps

1. find I.F = 
2. The solution will be y I.F = ∫ Q I.F dx + C

Formula used:

1. sin θ/cos θ = tan θ      2.1/cos θ = sec θ 

3. e^{ln x} = x       4. ∫ sec² x = tan x

Calculation:

cos x dy = y (sin x - y) dx

dy = y×(sin x - y) dx

⇒ dy = (y  - y²) dx ⇒  = y tan x - y² sec x 

⇒ 

Now, let y =  therefore 

Putting these values we get 

  ⇒ 

Now,  I.F = 

The solution of the equation will be 

⇒ t (I.F) =  ∫ (I.F) sec x dx + c ⇒ t (sec x) =  ∫ (I.F) sec x dx + c

⇒ t sec x = ∫ sec² x + c ⇒ sec x = (tan x + c)y

∴ The solution of an equation is sec x =  (tan x + c)y.

Concept:

In first order linear differential equation;

, where P and Q are function of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

IF = e∫  dx

⇒ IF = e^ln x

⇒ IF = x

Concept:

In the first-order linear differential equation;

,

Where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

x + y = 4x3 + x

⇒ 

IF = e∫  dx

⇒ IF = eln x

⇒ IF = x                 

(∵ e^{ln x} = x)

Now, y × (IF) = ∫ Q (IF) dx

⇒ y × x = ∫ (4x² + 1) × x dx

⇒ yx = ∫ 4x³ + x  dx

Integrating,

⇒ yx = x⁴ + + c (where c is integration constant)

⇒ y =