Solving Linear Differential Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Solving Linear Differential Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Form of a linear differential equation:

, where P and Q are the functions of x

Integrating factor = I.F = 

General solution is given by:

Calculation:

= 

Concept:

Linear differential equation:

A differential equation of the form  where P and Q are the functions of y or constants.

General solution of a linear differential equation:

The general solution of a linear differential equation of the form  is given by:

where I.F is known as Integrating Factor and it is calculated as follows: 

Calculation:

Express the given integral in the standard linear form of differential equation as:

Comparing with the standard form we get  and .

The integrating factor is calculated as:

Therefore, the general solution is given by:

Rearranging the terms, the general solution is given by .

Concept:

The solution of the linear differential equation  is given by

 y × I.F = 

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

Calculation:

Given    + y tan x = sec x 

This is a differential equation in the form 

Here P(x) = tan x and Q(x) = sec x

Integrating factor (I.F) = 

∴ I.F =  =  = sec x

The solution of the differential equation is given by:

y × I.F =  + C

⇒ y ⋅  sec x = 

⇒ y ⋅  sec x =  

∴ y ⋅  sec x =  tan x + c

The correct answer is option 1. 

Concept:

• The standard form of first order differential equation is  + Py = Q, where P and Q are constants or functions of x only.
• Integrating Factor (F) is given by: F = .

Calculation:

Let's first convert  into the standard form   + Py = Q.

∴ P = 

.

And, integrating factor F = .

Concept:

In the first-order linear differential equation;

, where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

General solution: y × (IF) = ∫ Q(IF) dx

Calculation:

So comparing the above equation with the linear differential equation

General solution:

At x  = 0 , y = 1

Therefore, 

Hence, option 3 is correct.

Concept:

The solution of the linear differential equation of the form  is given by

x × I.F = ∫I.F. × Q(y) dy + C

Here Integrating factor (I.F) = 

Calculation:

Given differential equation is (1 + y2) dx = (tan-1 y - x) dy

Rewriting the equation.

⇒ 

This is a differential equation in linear form 

Where P =  and Q = 

Integrating factor(I.F) =  

⇒ I.F = 

Solution of the differential equation is given by:

x × I.F =  + C

⇒ 

Put tan^-1y = t 

⇒ 

∴ 

⇒ 

⇒ 

⇒ x = (tan-1 y) - 1 +  

Required solution of the equation is x = (tan-1 y) - 1 + 

Concept Used:

The integrating factor (IF) for a first-order linear differential equation of the form is given by .

Calculation:

Given:

The differential equation is .

⇒

Here, and .

The integrating factor is:

⇒

⇒

Hence option 2 is correct

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  x + y = ex ⇒ 

This is a differential equation in the form 

Here P(x) =  and Q(x) = 

Integrating factor (I.F) =  =  = elog x = x

The solution of the differential equation is given by:

y × I.F =  ⇒ y ⋅ x = \(\int{x⋅ \frac{e^{x}}{x}}dx\)

⇒ y ⋅ x =  ⇒ y ⋅ x = ex + k

∴  y = 

The correct answer is option 1.

Concept:

In first-order linear differential equation;

,

Where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

General solution: y × (IF) = ∫ Q(IF) dx

Calculation:

Given the differential equation is

⇒ 

∴ It is linear differential equation is of first order

IF = e∫  dx

⇒ IF = e-ln x

⇒ IF = 

Now, y × (IF) = ∫ Q (IF) dx

⇒ y × = ∫ x ×  dx

⇒  = ∫ dx

Integrating,

⇒  = x + c (where c is integration constant)

Given y(1) = 1

⇒  = 1 + c

⇒ c = 0

∴   = x OR y = x² 

For y(2)

y = 22 

⇒ y = 4

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∫ sec² x dx

⇒ IF = e^tan x