Chapter 3 · Laplace Transforms

Solving ODEs with Laplace Transforms

Transform, Solve Algebraically, Invert

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Laplace Transform in Context

Definition & Properties ✓

Inverse Laplace ✓

Laplace Transform

Solving ODEs — THIS LECTURE!

Step Functions & Discontinuities

Convolution Theorem

The Payoff

      Solve any linear constant-coefficient ODE with initial conditions in just 3 steps. No guessing!

The 3-Step Method: The Big Idea

1

Transform

Apply Laplace transform to both sides of the ODE

\mathcal{L}\{f(t)\} = F(s)

2

Solve

Solve for $Y(s)$ algebraically (it's now an algebra problem!)

Y(s) = \frac{\text{algebraic expression}}{s}

3

Invert

Apply inverse Laplace to get $y(t)$

y(t) = \mathcal{L}^{-1}\{Y(s)\}

The Magic!

      Derivatives become polynomials in $s$! Initial conditions are automatically built in. No homogeneous/particular solution dance — just algebra.

Key Formulas: Transform of Derivatives

First-Order Derivative

      $$\mathcal{L}\{y'\} = sY(s) - y(0)$$
    
      The initial condition $y(0)$ is automatically included!

Second-Order Derivative

      $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$
    
      Both initial conditions $y(0)$ and $y'(0)$ appear automatically!

Why This Matters

      In classical methods, you find the general solution first, then apply ICs at the end. With Laplace, initial conditions are baked in from the start. This is a HUGE advantage!

The Complete 6-Step Method

STEP 1 Take $\mathcal{L}\{\}$ of both sides of the ODE

STEP 2 Use derivative transform formulas to substitute

STEP 3 Plug in known initial conditions

STEP 4 Collect all terms with $Y(s)$ on one side

STEP 5 Solve for $Y(s)$ — pure algebra!

STEP 6 Apply $\mathcal{L}^{-1}\{\}$ to find $y(t)$

Result

      You get the complete solution satisfying both the ODE and all initial conditions in one go!

The Transfer Function: System Characterization

Transfer Function Definition

      $$G(s) = \frac{1}{as^2 + bs + c}$$
    
      For a general linear ODE: $ay'' + by' + cy = f(t)$, the transfer function $G(s)$ captures the system's response characteristics. It depends only on the ODE itself, not on the forcing function or initial conditions.

Complete Solution Structure

      $$Y(s) = G(s) \cdot [F(s) + \text{IC terms}]$$
    
      The total response combines the forced response (from $F(s)$) and the natural response (from initial conditions).

Key Insight

      Once you know $G(s)$, you understand how the system responds to any input. This is why transfer functions are central to control theory and system analysis!

Worked Example 1

First-Order IVP: $y' + 2y = 6$, $y(0) = 1$

STEP 1 Take Laplace of both sides

\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{6\}

STEP 2 Apply derivative formula and IC

sY - 1 + 2Y = \frac{6}{s}

STEP 3 Collect $Y(s)$ terms and solve

(s + 2)Y = \frac{6}{s} + 1 = \frac{s + 6}{s}

Y = \frac{s + 6}{s(s + 2)}

STEP 4 Partial fractions

Y = \frac{3}{s} - \frac{2}{s + 2}

STEP 5 Invert using tables

y(t) = 3 - 2e^{-2t}

Worked Example 2

Second-Order Homogeneous: $y'' + 4y = 0$, $y(0) = 3$, $y'(0) = -2$

STEP 1-2 Take Laplace and apply formulas

$$s^2Y - 3s + 2 + 4Y = 0$$

STEP 3 Solve for $Y(s)$

$$(s^2 + 4)Y = 3s - 2$$

Y = \frac{3s}{s^2+4} - \frac{2}{s^2+4} = 3\frac{s}{s^2+4} - \frac{1}{2}\frac{2}{s^2+4}

STEP 4 Invert

y(t) = 3\cos(2t) - \sin(2t)

Worked Example 3

Driven System: $y'' + y = \sin(2t)$, $y(0) = 0$, $y'(0) = 0$

STEP 1-2 Take Laplace and apply

s^2Y + Y = \frac{2}{s^2+4}

STEP 3 Solve for $Y(s)$

Y = \frac{2}{(s^2+1)(s^2+4)}

STEP 4 Partial fractions with two quadratics

Y = \frac{2/3}{s^2+1} - \frac{2/3}{s^2+4}

STEP 5 Invert

y(t) = \frac{2}{3}\sin(t) - \frac{1}{3}\sin(2t)

Worked Example 4

Damped Oscillator: $y'' + 4y' + 3y = 0$, $y(0) = 1$, $y'(0) = 0$

STEP 1-2 Take Laplace and apply

$$s^2Y - s + 4(sY - 1) + 3Y = 0$$

STEP 3 Solve for $Y(s)$

$$(s^2 + 4s + 3)Y = s + 4$$

Y = \frac{s + 4}{(s+1)(s+3)}

STEP 4 Partial fractions

Y = \frac{3/2}{s+1} - \frac{1/2}{s+3}

STEP 5 Invert

y(t) = \frac{3}{2}e^{-t} - \frac{1}{2}e^{-3t}

Worked Example 5

Exponential Forcing: $y'' - y = e^{2t}$, $y(0) = 0$, $y'(0) = 1$

STEP 1-2 Take Laplace and apply

s^2Y - 1 - Y = \frac{1}{s-2}

STEP 3 Solve for $Y(s)$

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

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

STEP 4 Partial fractions

Y = -\frac{1/3}{s+1} + \frac{1/3}{s-2}

STEP 5 Invert

y(t) = -\frac{1}{3}e^{-t} + \frac{1}{3}e^{2t}

Worked Example 6

Repeated Roots: $y'' - 2y' + y = 0$, $y(0) = 1$, $y'(0) = 3$

STEP 1-2 Take Laplace and apply

$$s^2Y - s - 3 - 2(sY - 1) + Y = 0$$

STEP 3 Solve for $Y(s)$

$$(s^2 - 2s + 1)Y = s + 1$$

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

STEP 4 Partial fractions (repeated root)

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

STEP 5 Invert (use $\mathcal{L}^{-1}\{\frac{1}{(s-a)^2}\} = te^{at}$)

y(t) = e^{t} + 2te^{t} = (1+2t)e^{t}

Worked Example 7

Complex Roots + Constant Forcing: $y'' + 2y' + 5y = 10$, $y(0) = 0$, $y'(0) = 0$

STEP 1-2 Take Laplace and apply

s^2Y + 2sY + 5Y = \frac{10}{s}

STEP 3 Solve for $Y(s)$

Y = \frac{10}{s(s^2+2s+5)}

STEP 4 Partial fractions: $\frac{A}{s} + \frac{Bs+C}{s^2+2s+5}$ where $s^2+2s+5 = (s+1)^2+4$

A = 2, \quad B = -2, \quad C = -4

STEP 5 Invert

y(t) = 2 - 2e^{-t}\cos(2t) - e^{-t}\sin(2t)

Real-World Application

RLC Circuit with Constant Voltage

RLC Circuit Equation

      $$LQ'' + RQ' + \frac{Q}{C} = V(t)$$
    
      where $Q(t)$ is charge, $L$ is inductance, $R$ is resistance, $C$ is capacitance, $V(t)$ is applied voltage.

Numerical Example

Given: $L=1$ H, $R=4$ Ω, $C=1/3$ F, $V=10$ V (constant), $Q(0)=0$, $Q'(0)=0$

$$Q'' + 4Q' + 3Q = 10$$

Laplace transform gives:

Y = \frac{10}{s(s^2+4s+3)} = \frac{10}{s(s+1)(s+3)}

After partial fractions and inversion:

Q(t) = \frac{10}{3} - 5e^{-t} + \frac{5}{3}e^{-3t}

Physical Interpretation

      The charge exponentially approaches $10/3$ coulombs (steady state) from below, with the transient decaying as two exponential terms.

Laplace Transform vs Classical Methods

Classical Method

          1. Guess solution form

          2. Solve auxiliary/characteristic equation

          3. Find homogeneous solution

          4. Find particular solution (undetermined coefficients, variation of parameters)

          5. Apply initial conditions at the end to find constants

Laplace Method

          1. Transform both sides

          2. Solve algebraically for $Y(s)$ (ICs built in!)

          3. Use partial fractions

          4. Invert to get $y(t)$

          Done! The complete solution with all ICs satisfied.

When to Use Each

      Classical: Fast for simple equations, good for gaining intuition.

      Laplace: Preferred for systems with ICs, discontinuous forcing (step functions), convolution, and complex systems. Works for all linear constant-coefficient ODEs.

Common Pitfalls & Mistakes

⚠️ Mistakes to Avoid

Forgetting IC terms in derivative transforms: $\mathcal{L}\{y''\}$ MUST include both $y(0)$ and $y'(0)$ terms. Don't skip them!
Algebra errors solving for $Y(s)$: Keep track of signs carefully. It's easy to drop a minus sign and get the wrong answer.
Wrong partial fraction form: For quadratic denominators like $(s^2+1)$, use $\frac{As+B}{s^2+1}$, not $\frac{A}{s^2+1}$.
Not simplifying before inversion: Always cancel common factors and simplify $Y(s)$ completely before inverting.
Mixing up the inverse transforms: Memorize: $\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at}$ and $\mathcal{L}^{-1}\{\frac{1}{(s-a)^2}\} = te^{at}$, etc.
Forgetting to check your answer: If time permits, verify: does $y(t)$ satisfy the ODE? Do the ICs check out?

Key Takeaways & Summary

The 3-Step Workflow

1. Transform both sides of the ODE
2. Solve for $Y(s)$ algebraically
3. Invert to get $y(t)$

Derivative Formulas

$\mathcal{L}\{y'\} = sY(s) - y(0)$
$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$
ICs are built in!

Transfer Function

$G(s) = \frac{1}{\text{characteristic polynomial}}$
Captures system response independently of input/ICs

Works for All Linear Constant-Coeff ODEs

First-order, second-order, higher-order
Homogeneous, non-homogeneous
Any forcing function

Next Up

      Step Functions & Convolution! What happens when the forcing function is discontinuous (switches on/off)? Laplace handles this elegantly using the Heaviside step function and convolution theorem.