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Test: Method of Separation of Variables - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Method of Separation of Variables

Test: Method of Separation of Variables for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Method of Separation of Variables questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Method of Separation of Variables MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Method of Separation of Variables below.
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Test: Method of Separation of Variables - Question 1

The solution of  dy/dx = y with initial value y(0) = 1 is bounded in the interval 

Detailed Solution for Test: Method of Separation of Variables - Question 1


⇒ from the Integrating the equation, we get

⇒ at x = 0, y = 1, hence -1/1 = c ⇒ c = -1

x ≠ 1, at (x = 1) y will be undefined

x > 1, x < 1

Test: Method of Separation of Variables - Question 2

Solve the following equation:
yexydx + (xexy + 2y)dy = 0

Detailed Solution for Test: Method of Separation of Variables - Question 2

The given equation is in the form of

M dx + N dy = 0

Here M = y exy

N = (x exy + 2y)

 

Hence the differential equation is an exact equation.

The solution is

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Test: Method of Separation of Variables - Question 3

Consider the differential equationThe general solution with constant C is

Detailed Solution for Test: Method of Separation of Variables - Question 3

Concept:

Separation of Variables:

Calculation:

Given:

Differential equation:

By separation of variables:

Integrating on both sides we will get:

Test: Method of Separation of Variables - Question 4

The solutions of the equation 3yy’ + 4x = 0 represents a:

Detailed Solution for Test: Method of Separation of Variables - Question 4

3yy’ + 4x = 0

3ydy = - 4x dx

On integration both the sides:

Where c is the constant of integration.

Thus, the solution of given differential equation represents the family of Ellipses.

Test: Method of Separation of Variables - Question 5

The solution of differential equation dx – (x + y + 1) dy = 0 is

Detailed Solution for Test: Method of Separation of Variables - Question 5

Given differential equation is,

dx – (x + y + 1) dy = 0

Put (x + y + 1) = t

Now, the differential equation becomes


⇒ t - In (t + 1) = x + C1

⇒ (x + y + 1) – In (x + y + 2) = x + C1

⇒ y + 1 – C1 = In (x + y + 2)

⇒ In (x + y + 2) = y + k

⇒ (x + y + 2) = ey+k

⇒ (x + y + 2) e-y = C

Test: Method of Separation of Variables - Question 6

The solution of the differential equation (dy/dx) = ky, y(0) = c is

Detailed Solution for Test: Method of Separation of Variables - Question 6

The given differential equation is, (dy/dx) = ky

dy/y = kdx

On integrating both the sides, we get

ln y = kx + ln c

y = cekx

Test: Method of Separation of Variables - Question 7

The solution of the differential equation 

Detailed Solution for Test: Method of Separation of Variables - Question 7

Given differential equation is,


By integrating both sides,

Test: Method of Separation of Variables - Question 8

What is the solution of the differential equation dy/dx = x/y, with the initial condition, x = 0, y = 1?

Detailed Solution for Test: Method of Separation of Variables - Question 8

dy/dx = x/y

⇒ ydy = xdx

By integrating on both sides, we get

Given the initial condition, x = 0, y = 1.

⇒ 1/2 = 0 + C

⇒ C = 1/2

Now, the solution becomes

y2 = x2 + 1

*Answer can only contain numeric values
Test: Method of Separation of Variables - Question 9

Given  the value of y at x = 2 is ________ (round off to nearest integer)      (Important - Enter only the numerical value in the answer)


Detailed Solution for Test: Method of Separation of Variables - Question 9

Given the differential equation is,


By integrating with respect to ‘x’, we get

⇒ ln(20) = C

Now, the differential equation becomes

At x = 2, y = 20 (e2 + 1) = 167.78

Test: Method of Separation of Variables - Question 10

Which of the following equations cannot be solved by using the method of separation of variables?

Detailed Solution for Test: Method of Separation of Variables - Question 10

The method of separation of variables is used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as:

  • Heat equation
  • Wave equation
  • Laplace equation
  • Helmholtz equation
  • Biharmonic equation
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