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Test: Method of Parameter Variations - Civil Engineering (CE) MCQ


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9 Questions MCQ Test Engineering Mathematics - Test: Method of Parameter Variations

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

Solve (x2 – y2) dx – xydy = 0

Detailed Solution for Test: Method of Parameter Variations - Question 1

(x2 – y2) dx – xydy = 0

This homogeneous in x & y

Separating the variables

Integrating on both side

⇒ 4logx + log⁡(1 − 2v2) = −4c

⇒ x2 (x2 – 2y2) = constant

Test: Method of Parameter Variations - Question 2

The differential equationis solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)

Detailed Solution for Test: Method of Parameter Variations - Question 2

Concept:

Calculation:

Given:

(D2 – 6D + 9)y = e3x/x2

Auxiliary equation is

(D2 – 6D + 9) = 0

⇒ (D – 3)2 = 0

C.F. = (C1 + C2x) e3x

C.F = C1(e3x) + C2(xe3x)

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Test: Method of Parameter Variations - Question 3

The differential equation is solving by the method of variation of parameters, the Wronskian will be______

Detailed Solution for Test: Method of Parameter Variations - Question 3


Auxiliary equation, D2 – 1 = 0

⇒ D = ± 1

C.F. = C1ex + C2e−x

Here, y1 = ex

y2 = e-x

y1′ = ex
y2′ = −e−x


= -1 - 1 = -2

Test: Method of Parameter Variations - Question 4

The Wronskian of the differential equation

Detailed Solution for Test: Method of Parameter Variations - Question 4

If the given differential equation is in the form, y'' + py' + qy = x

Where, p, q, x are functions of x.

Then Wronskian is,

Here, y1 and y2 are the solutions of y'' + py' + qy = 0

Given differential equation is,

Auxiliary equation is,

D2 – 3D + 2 = 0

⇒ D = 1, 2

C.F. = c1 e2x + cex

y1 = e2x, y2 = ex

Test: Method of Parameter Variations - Question 5

Consider the following differential equation:

Which of the following is the solution of the above equation (is an arbitrary constant)?

Detailed Solution for Test: Method of Parameter Variations - Question 5

Given differential eqaution is,

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

Integrating both sides

2 log x = log |sec v| - log v + log c

Test: Method of Parameter Variations - Question 6

The differential equationis solving by the method of variation of parameters, then Wronskian will be –

Wronskian for solution y = c1y1(t) + c2y2(t) is defined as

Detailed Solution for Test: Method of Parameter Variations - Question 6

Give equation is (D2 – 6D + 9)y = e3x/x2

Auxiliary equation is, (D2 – 6D + 9) = 0

⇒ (D – 3)2 = 0

C.F. = (C1 + C2x) e3x

Test: Method of Parameter Variations - Question 7

Find the particular solution of the differential equation

Detailed Solution for Test: Method of Parameter Variations - Question 7

Given differential equation is

auxiliary equation is (D − 2)2 = 0 ⇒ D = 2, 2


This problem can be solved by using the method of variation of parameters. Then

Test: Method of Parameter Variations - Question 8

Solution of the differential equation is 

Detailed Solution for Test: Method of Parameter Variations - Question 8


Test: Method of Parameter Variations - Question 9

 

Solution of differential equation (D2 + 4)y = cosec 2x

Detailed Solution for Test: Method of Parameter Variations - Question 9

Auxiliary equation:

m2 + 4 = 0

m = ±2i


By wrongskion,

PI = A u(x) + B v(x)

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