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Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mathematics - Test: Higher Order Linear Differential Equations with Constant Coefficients - 1

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

A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacement of the system at any time t, the equation of motion for the free vibration of the system is The natural frequency of the system is

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

Comparing with(Standard Differential Equation for Spring)

Natural frequency,

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

The solution of the differential equation, for t > 0, y″(t) + 2y′(t) + y(t) = 0 with initial conditions y′(0) = 1and y(0) = 0, is u(t) denotes the unit step functions).

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

Concept:

 Laplace transform of the first two derivatives.

L{y''} = s2Y(s) - sy(0) - y'(0)

L{y'} = sY(s) - y(0)

Calculation:

Given y″(t) + 2y′(t) + y(t) = 0

Apply Laplace transform we get

[s2Y(s) − sy(0) − y′(0)] + 2[sy(s) − y(0)] + Y(s) = 0
(s2 + 2s + 1)Y(s)

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Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 3

The solution to the differential equationwhere k is a constant, subjected to the boundary conditions u(0) = 0 and u(L) = U, is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 3

Concept:

If roots are real and different then

Calculation:

Given:

D2 – kD = 0    

m(m - k) = 0

m1 = 0, m2 = k 

Roots are real and different

u(0) = 0

∴ C1 + C2 = 0  .................... (2)

u(L) = U

solving (2)  and (3), we get


Putting values fo C1 and C2 in eqn 1

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 4

A function y(t) such that y(0) = 1 and y(1) = 3e-1, is a solution of the differential equation . Then y(2) is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 4


By applying the Laplace transform,
s2Y(s) − sy(0) − y′(0) + 2sY(s) − 2y(0) + Y(s) = 0

By applying the inverse Laplace transform

y(t) = e−t + (1 + y′(0))te−t

y(1) = 3 e-1

⇒ y(1) = e-1 + (1 + y’(0)) e-1 = 3 e-1

⇒ y’(0) = 1

Now the equation of y(t) becomes

y(t) = (1 + 2t) e-t

At t = 2,

y(2) = 5 e-2

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 5

Consider the following statements about the linear dependence of the real valued functions y1 = 1, y2 = x and y3 = x2, over the field of real numbers.

I. y1, y2 and y3 are linearly independent on – 1 ≤ x ≤ 0

II. y1, y2 and y3 are linearly dependent on 0 ≤ x ≤ 1

III. y1, y2 and y3 are linearly independent on 0 ≤ x ≤ 1

IV. y1, y2 and y3 are linearly dependent on – 1 ≤ x ≤ 0

Which one among the following is correct?

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 5

Concept:

The linear dependency or Independency of function fi can be found using Wronskion Matrix.

An analytic function is linearly dependent if:

Application:

Determinant ≠ 0

Function y1, y2, y­3 are linearly independent.

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 6

For the Ordinary Differential Equationwith initial conditions x(0) = 0 and dx/dt(0) = 10, the solution is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 6

Above given equation is a linear differential equation of order = 2. Such equations are solved using CF + PI method.

Let D = dx/dt


⇒ D2x – 5Dx + 6x = 0

⇒ (D2 – 5D + 6)x = 0

Thus auxiliary equation (obtained by replacing D with m) is m2 – 5D + 6 = 0.

Roots of above obtained auxiliary equation are-

m2 – 5D + 6 = 0

⇒ (m – 2) (m - 3) = 0

⇒ m1 = 2 or m2 = 3

When both roots of auxiliary equation are real and distinct

Thus general solution of above D.E is

Applying boundary conditions, x(0) = 0

⇒ O = C1 + C2

⇒ C1 = -C2     ---(i)

Using initial condition dx/dt(0) = 10

⇒ 10 = 2C1 + 3C2       ---(ii)

Solving equation (i) & (ii) simultaneously,

C1 = -10 and C2 = 10

Thus, x = -10e2t + 10e3t

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 7

If roots of the auxiliary equation ofare real and equal, the general solution of the differential equation is

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 7

Concept:

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

Calculation:

Given:

⇒ (D2 + aD + b)y = 0

Auxiliary equation:

m2 + am + b = 0

Roots of the auxiliary equation are:

Given that, the roots of the auxiliary equation are real and equal.

⇒ m = -a/2   [∵ a2 - 4ab = 0]

The general solution of the differential equation is: 

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 8

The general solution of the differential equation in terms of arbitrary constants K1 and K2 is:

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 8

Given Differential equation:

Auxillary equation = D2 + 2D – 5 = 0

Solving, we get roots

Therefore the general solution is;

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 9

The complete solution of the linear differential equation

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 9

For the given differential equation:


(D2 + (p + q)D + pq)z = 0

⇒ f(D)z = 0

Where f(D) = D2 + (p + q)D + pq

Consider in terms of M ⇒ f(m) = 0

M2 + (p + q)m + pq = 0


M1 = -q & m2 = -p

Hence, its solution is

Z = C1e-pt + C2e-qt

∴ Complete solution of the linear differential equation is ce-pt + ce-qt

Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 10

The differential equationFor y(x) with the two boundary conditions

Detailed Solution for Test: Higher Order Linear Differential Equations with Constant Coefficients - 1 - Question 10

Concept:

Given equation is

This is a homogeneous second order differential equation,

So (D2 + 16)y = 0

D2 = m2

⇒ m2 + 16 = 0     ⇒ m = ± 4i = 0 ± 4i

Solution is given as in this case roots are complex, m = α ± i β

y = (C1 cos βx + C2 sin βx) eαx

= (C1 cos 4x + C2 sin 4x) eox = C1 cos 4x + C2 sin 4x

Now y’ = -4C1 sin 4x + 4C2 cos 4x

Applying Boundary condition,

y’ (0) = 1  ⇒ -4C1 sin (0) + 4C2 cos(0) = 1

4C2 = 1 ⇒ C2 = 1/4

Putting another boundary condition.

y′(π/2) = −1


So this equation has no solution.

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