Introduction

The Laplace Transform converts a function of time \(f(t)\) into a function of a complex variable \(s\):

\[ \mathcal{L}\{f(t)\}=F(s)=\int_{0}^{\infty} f(t) e^{-st} \, dt \]

Why Use Laplace Transforms?

  • Convert differential equations into algebraic equations
  • Automatically incorporate initial conditions
  • Handle discontinuous forcing functions (step, impulse)
  • Provide systematic solution method for linear ODEs

Common Laplace Transforms

\(f(t)\) for \(t \ge 0\) \(F(s)\) Condition
\(\delta(t)\) (Dirac delta) \(1\)
\(1\) (unit step) \(\frac{1}{s}\) \(s > 0\)
\(t^n\) \(\frac{n!}{s^{n+1}}\) \(s > 0\)
\(e^{at}\) \(\frac{1}{s-a}\) \(s > a\)
\(\sin(\omega t)\) \(\frac{\omega}{s^2 + \omega^2}\) \(s > 0\)
\(\cos(\omega t)\) \(\frac{s}{s^2 + \omega^2}\) \(s > 0\)
\(e^{at} \sin(\omega t)\) \(\frac{\omega}{(s-a)^2 + \omega^2}\) \(s > a\)
\(e^{at} \cos(\omega t)\) \(\frac{s-a}{(s-a)^2 + \omega^2}\) \(s > a\)

Key Properties

Operation Time Domain Laplace Domain
Linearity \(af(t)+bg(t)\) \(aF(s)+bG(s)\)
First derivative \(f'(t)\) \(sF(s)-f(0)\)
Second derivative \(f''(t)\) \(s^2F(s)-sf(0)-f'(0)\)
\(n\)-th derivative \(f^{(n)}(t)\) \(s^nF(s)-\sum_{k=1}^{n} s^{n-k} f^{(k-1)}(0)\)
Integration \(\int_{0}^{t} f(\tau) d\tau\) \(\frac{F(s)}{s}\)
\(t\)-multiplication \(tf(t)\) \(-F'(s)\)
Frequency shift \(e^{at}f(t)\) \(F(s-a)\)
Time shift (for \(t \ge 0\)) \(f(t-a)u(t-a)\) \(e^{-as}F(s)\)

Solving Differential Equations: Step-by-Step

General Procedure

  1. Take Laplace transform of both sides of the ODE
  2. Use initial conditions to simplify
  3. Solve the algebraic equation for \(Y(s)\)
  4. Perform partial fraction decomposition
  5. Use inverse Laplace transform to find \(y(t)\)

Example: Second-Order Homogeneous ODE

Problem: Solve \(y'' + 3y' + 2y = 0\) with \(y(0)=1\), \(y'(0)=0\)

Step 1: Take Laplace transform

\[ \mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = 0 \]

\[ [s^2 Y(s) - s y(0) - y'(0)] + 3[s Y(s) - y(0)] + 2Y(s) = 0 \]

Step 2: Substitute initial conditions \(y(0)=1\), \(y'(0)=0\):

\[ s^2 Y(s) - s + 3s Y(s) - 3 + 2Y(s) = 0 \]

Step 3: Solve for \(Y(s)\)

\[ (s^2 + 3s + 2)Y(s) = s + 3 \]

\[ Y(s) = \frac{s+3}{s^2 + 3s + 2} = \frac{s+3}{(s+1)(s+2)} \]

Step 4: Partial fractions

\[ \frac{s+3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \]

\[ s+3 = A(s+2) + B(s+1) = (A+B)s + (2A+B) \]

Comparing coefficients: \(A+B=1\), \(2A+B=3\)

Solving: \(A=2\), \(B=-1\)

Thus: \(Y(s) = \frac{2}{s+1} - \frac{1}{s+2}\)

Step 5: Inverse Laplace transform

\[ y(t) = 2e^{-t} - e^{-2t} \]

Verification using R

y <- function(t) {
  2*exp(-t) - exp(-2*t)
}

cat("y(0) =", y(0), "\n")
## y(0) = 1
y_prime <- function(t) {
  -2*exp(-t) + 2*exp(-2*t)
}
cat("y'(0) =", y_prime(0), "\n")
## y'(0) = 0
t_vals <- seq(0, 5, length.out = 100)
y_vals <- y(t_vals)

plot(t_vals, y_vals, type = "l", lwd = 2, col = "blue",
     xlab = "t", ylab = "y(t)", main = "Solution: y(t) = 2e^{-t} - e^{-2t}")
grid()