Test: Higher Order Linear Differential Equations with Constant Coefficients - 2 - Computer Science Engineering (CSE) MCQ

Given:

(D + 3)²y = 5^x - log 2

Formula Used:

e ^{log x} = x

Calculation:

We have,

⇒ (D + 3)²y = 5^x - log 2

Particular Integral of the differential equation is


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

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 a² in 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:

D² + 25x = 21 × Sin (t)

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

∴ The displacement at any time t > 0 = 

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

= e^mx

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 e^t + c₂ e^-t

y(x) = c₁ x + c₂ x^-1

Concept:

For solving a homogeneous linear differential equation with constant coefficients, 

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

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

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

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

Calculation:

Given Differential equation is:

u_n+3 - 4_n+2 + u_n+1 + 6u_n = 0

Substituting:

u_n+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:

u_n = A(-1)ⁿ + B(2)ⁿ + C(3)ⁿ

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:

D³ y - 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 = y_c = c₁e^m₁x + c₂e^m₂x + c₃e^m₃x

y_c = c₁e^x + c₂e^-x + c₃e^2x

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

Given differential equation:

is the standard form of Euler-Cauchy DE.

So, let x = e^z

⇒ dx = e^zdz = 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 = c₁e^z + c₂e^2z = c₁x + c₂x² 

⇒ y = CF + PI = c₂x² + c₁x + 2

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

Concept:

When Φ(x) = e^ax + V(x)

If f(a) = 0

Calculation:

Concept:

General equation for DE:

Then its corresponding Auxiliary equation will be

AE: Dⁿ + k₁ D^n-1 + … k_n = 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) e^x + c₃ cos x + c₄ sin x = 0

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.