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

Comparing with(Standard Differential Equation for Spring)

Natural frequency,

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)

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

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

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.

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 = -10e^2t + 10e^3t

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: 

Given Differential equation:

Auxillary equation = D² + 2D – 5 = 0

Solving, we get roots

Therefore the general solution is;

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 c₁ e^-pt + c₂ e^-qt

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) e^αx

= (C₁ cos 4x + C₂ sin 4x) e^ox = 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.