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Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mathematics - Test: Higher Order Linear Differential Equations with Constant Coefficients - 2

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Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 1

The particular integral of (D + 3)2 y = 5x - log 2, is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 1

Given:

(D + 3)2y = 5x - log 2

Formula Used:


log x = x

Calculation:

We have,

⇒ (D + 3)2y = 5x - log 2

Particular Integral of the differential equation is


∴ The particular integral of (D + 3)2y = 5x - log 2 is 

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 2

A particle undergoes forced vibrations according to the law xn(t) + 25x(t) = 21 sin t. If the particle starts from rest at t = 0, find the displacement at any time t > 0. 

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 2

Concept:

The solution of the linear differential equation is of the form:

y = C.F + P.I

where C.F is the complementary function and P.I is the particular integral.

The 2nd order linear differential equation in the symbolic form is represented as:

When X = Sin(ax)

f(-a2) is calculated by replacing D2 with ain f(D)

In the case of a forced damped system, the vibration equation:

 The solution of x is (CF + PI)

Where CF is complimentary function and PI is particular integral

But after some time CF becomes zero.

∴ x = P.I

Calculation:

Given:

xn(t) + 25x(t) = 21 sin t

The above equation can be written as:

D+ 25x = 21 × Sin (t)

Here, a = 1, So putting -a2 = D2 = -1 in the above equation

∴ The displacement at any time t > 0 = 

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Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 3

An integrating factor for the differential equation 

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 3

Concept:

If a first-order linear differential equation can be expressed as,

where P and Q are either function of x or constant

And integrating factor (I.F.) is given by

I.F = e∫Pdx

Then, The solution of the differential equation is given by,

y × e∫Pdx = ∫Q × e∫Pdxdx + c

where c is an integrating constant.

Calculation:

Given:

The given differential equation,

Here, P = m, Q = e−mx

I.F = e∫Pdx

= e∫mdx

= emx

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 4

The solution to x2y’’ + xy’ – y = 0 is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 4

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

Given:

Put x = et

⇒ t = ln x

Now, the above differential equation becomes

D(D – 1)y + Dy - y = 0

⇒ D2y – Dy + Dy - y = 0

⇒ (D2 -1)y = 0

Auxiliary equation:

(m2 – 1) = 0

⇒ m = ±1

The solutions for the above roots of auxiliary equations are: 

y(t) = c1 et + c2 e-t

y(x) = c1 x + c2 x-1

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 5

The solution of differential equation un+3 - 4un+2 + un+1 + 6un = 0 will be:

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 5

Concept:

For solving a homogeneous linear differential equation with constant coefficients, 

As a solution, we try un = A mn, where A and m are constants.

This choice has been made because un = kn uo was the solution to the equation un = k un-1.

After substituting and solving the auxiliary equation, say m1, m2, m3 ... mn be the roots, then the general solution will be 

un = A(m1)n + B(m2)n + ... where A, B ... are constants

Calculation:

Given Differential equation is:

un+3 - 4n+2 + un+1 + 6un = 0

Substituting:

un+3 = A mn+3

un+2 = A mn+2; 

un+1 = A mn+1; 

un = A mn; 

A mn+3 - 4 A mn+2 + A mn+1 + 6 A mn = 0

A mn (m3 - 4m2 + m + 6) = 0

m3 - 4m2 + m + 6 = 0

This is called auxiliary equation.

The roots of the auxiliary equation are -1, 2, and 3, i.e.

m1 = -1, m2 = 2, m3 = 3;

The general solution will be:

un = A(-1)n + B(2)n + C(3)n

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 6

The solution ex, e-x, and e2x of will be:

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 6

Concept:

Write auxillary equation by replacing

Equate to 0 and solve for (m) (f(m) = 0)

for m1, m2, m3, … (real and different roots)

Complementary function (CF) is given as:

Calculation:


The above expression is in the form f(D)y = X and can be written as:

Dy - 2D2y - Dy + 2y = 0

y(D3 - 2 × D2 - D + 2) = 0

D3 - 2D2 - D + 2 = 0

The auxiliary equation is f(m) = 0

m3 - 2m2 - m + 2 = 0

(m - 1)(m + 1)(m - 2) = 0

So, the roots the auxiliary equation are m1 = 1, m2 = -1, m3 = 2.

The characteristics of the roots are real and distinct.

C.F = yc = c1em1x + c2em2x + c3em3x

yc = c1e+ c2e-x + c3e2x

So the solution will be linearly independent on every real interval.

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 7

A differential equation is given as:

The solution of the differential equation in terms of arbitrary constants Cand C2 is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 7

Given differential equation:

is the standard form of Euler-Cauchy DE.

So, let x = ez

⇒ dx = ezdz = x.dz

Similarly, we can obtain

⇒ xD = θ

⇒ x2D2 = θ (θ - 1)

Making these substitutions in given DE, we get

(θ (θ - 1) – 2θ + 2) y = 4

θ (θ - 1) – 2(θ – 1) = 0

⇒ Solution for this will constitute of CF & PI.

⇒ CF = (θ – 1) (θ – 2) = 0 ⇒ θ = 1 & θ = 2

⇒ y = c1ez + c2e2z = c1x + c2x

⇒ y = CF + PI = c2x2 + c1x + 2

{We can also solve this problem by differentiating option also, if we don’t remember the process.}

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 8

Particular integral of 

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 8

Concept:

When Φ(x) = eax + V(x)

If f(a) = 0

Calculation:


Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 9

The general solution of the differential equation 

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 9

Concept:

General equation for DE:

Then its corresponding Auxiliary equation will be

AE: Dn + k1 Dn-1 + … kn = 0

Then the solution of above DE will be as follows:

Calculation:

Given:

Its Auxillary equations:

(D4 – 2D3 + 2D2 – 2D + D) = 0

(D - 1)(D3 - D2 + D - 1) = 0

(D - 1)(D - 1)(D2 + 1) = 0

It roots = 1, 1, i, -i

Then it has two equal roots and one pair of imaginary roots.

Therefore, the solution is:

(c1 + c2x) ex + c3 cos x + c4 sin x = 0

Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 10

The ordinary differential equation

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Question 10

Concept:

Identification of Non-linear Differential Equation:

If any differential equation consists at least one of the above properties, then it is called non-linear differential equation and if any differential equation is free from all the above properties, then it is a linear differential equation.

Given:

It is free from all the four characteristics described in the table. Hence it is a linear differential equation.

Product of dependent and independent variable i.e. x and u is present. Hence it is a non-homogenous equation.

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