Math Problem Statement

Determine the Laplace transforms of given functions and solve differential equations using Laplace transforms.

Solution

Let’s break this question paper into solutions for each problem step by step.


QUESTION TWO

Part (a): Laplace Transform

  1. (i) 5s+1s2s12\frac{5s+1}{s^2 - s - 12}

    • To find the inverse Laplace transform of a rational function, we typically factorize the denominator s2s12s^2 - s - 12.
    • Factorize: s2s12=(s4)(s+3)s^2 - s - 12 = (s - 4)(s + 3)
    • Rewrite the function using partial fraction decomposition: 5s+1(s4)(s+3)=As4+Bs+3\frac{5s+1}{(s-4)(s+3)} = \frac{A}{s-4} + \frac{B}{s+3} Multiply through: 5s+1=A(s+3)+B(s4)5s + 1 = A(s+3) + B(s-4) Expand and solve for AA and BB.
  2. (ii) 3s+1s2s6\frac{3s+1}{s^2 - s - 6}

    • Factorize the denominator: s2s6=(s3)(s+2)s^2 - s - 6 = (s - 3)(s + 2)
    • Apply partial fraction decomposition similarly: 3s+1(s3)(s+2)=Cs3+Ds+2\frac{3s+1}{(s-3)(s+2)} = \frac{C}{s-3} + \frac{D}{s+2} Solve for constants CC and DD to complete.

Part (b): Solve Differential Equation Using Laplace Transform

The given equation is: dydx2y=4,y(0)=1\frac{dy}{dx} - 2y = 4, \quad y(0) = 1 Steps to solve:

  1. Take the Laplace Transform of both sides:

    • Recall properties: L(dydx)=sY(s)y(0),L(y)=Y(s)\mathcal{L}\left(\frac{dy}{dx}\right) = sY(s) - y(0), \quad \mathcal{L}(y) = Y(s)
  2. Substitute into the equation: sY(s)y(0)2Y(s)=L(4)=4ssY(s) - y(0) - 2Y(s) = \mathcal{L}(4) = \frac{4}{s} Simplify and solve for Y(s)Y(s).

  3. Apply the Inverse Laplace Transform to find y(t)y(t).


QUESTION THREE

Part (b): Define Periodic Function

The given periodic function f(x)f(x) has a repeating triangular waveform with peaks and zero crossings:

  • Period T=6T = 6 (based on the graph).
  • Define analytically using a piecewise function:
    • f(x)=332xfor 0x<3f(x) = 3 - \frac{3}{2}x \quad \text{for } 0 \leq x < 3
    • f(x)=3+32xfor 3x<6f(x) = -3 + \frac{3}{2}x \quad \text{for } 3 \leq x < 6
  • Extend periodically with f(x+6)=f(x)f(x + 6) = f(x).

Part (c): Solve Differential Equation with Laplace Transform

The equation is: d2xdt23dxdt+2x=2e3t,x(0)=5,dxdtt=0=7\frac{d^2x}{dt^2} - 3\frac{dx}{dt} + 2x = 2e^{3t}, \quad x(0) = 5, \, \frac{dx}{dt}\Big|_{t=0} = 7 Steps to solve:

  1. Take the Laplace Transform of both sides:
    • Use: L(d2xdt2)=s2X(s)sx(0)x(0),L(dxdt)=sX(s)x(0)\mathcal{L}\left(\frac{d^2x}{dt^2}\right) = s^2X(s) - sx(0) - x'(0), \quad \mathcal{L}\left(\frac{dx}{dt}\right) = sX(s) - x(0)
  2. Substitute initial conditions x(0)=5x(0) = 5 and x(0)=7x'(0) = 7.
  3. Solve for X(s)X(s) (Laplace domain).
  4. Apply the Inverse Laplace Transform to find x(t)x(t).

Questions for Deeper Understanding:

  1. How do we factorize higher-order polynomials when solving Laplace transforms?
  2. What is the importance of initial conditions in solving differential equations?
  3. Can you explain the periodic function using Fourier series representation?
  4. How do Laplace transforms handle non-homogeneous equations?
  5. What happens if the periodic function has a non-zero average value?

Tip:

When using Laplace Transforms to solve ODEs, always remember to carefully apply initial conditions after transforming the equation.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Differential Equations
Partial Fractions
Periodic Functions

Formulas

Laplace Transform of derivative: L{y'} = sY(s) - y(0)
Inverse Laplace Transform
Partial Fraction Decomposition for rational functions

Theorems

Laplace Transform properties
Superposition Principle for solving differential equations

Suitable Grade Level

University Level