Lesson 2: Inverse Laplace and Partial Fractions
Estimated time: 45-55 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Use the Laplace transform table in reverse to find inverse transforms
- Decompose rational functions using partial fractions (distinct linear, repeated, irreducible quadratic)
- Complete the square to handle quadratic denominators
- Apply linearity of the inverse Laplace transform
- Combine techniques to invert complex rational expressions
The Inverse Laplace Transform
Inverse Laplace Transform: If F(s) = L{f(t)}, then f(t) = L-1{F(s)}. The inverse is linear: L-1{αF + βG} = αf + βg. We use partial fractions to break F(s) into forms matching our table.
Partial Fractions: Distinct Linear Factors
Distinct Linear Factors: If the denominator factors as (s-a)(s-b) with a ≠ b, write F(s) = A/(s-a) + B/(s-b). Then L-1{F} = Aeat + Bebt.
Example 1: Distinct Linear Factors
Find L-1{(3s+1)/[(s-1)(s+2)]}.
Step 1: Write (3s+1)/[(s-1)(s+2)] = A/(s-1) + B/(s+2).
Step 2: Multiply through: 3s+1 = A(s+2) + B(s-1).
Step 3: Set s=1: 4 = 3A, A = 4/3. Set s=-2: -5 = -3B, B = 5/3.
Answer: L-1 = (4/3)et + (5/3)e-2t.
Example 2: Three Distinct Factors
Find L-1{6/[s(s+1)(s+3)]}.
Step 1: 6/[s(s+1)(s+3)] = A/s + B/(s+1) + C/(s+3).
Step 2: 6 = A(s+1)(s+3) + Bs(s+3) + Cs(s+1).
s=0: 6=3A, A=2. s=-1: 6=-2B, B=-3. s=-3: 6=6C, C=1.
Answer: L-1 = 2 - 3e-t + e-3t.
Partial Fractions: Repeated Linear Factors
Repeated Factor: For (s-a)k in the denominator, include A1/(s-a) + A2/(s-a)² + ... + Ak/(s-a)k. Key formula: L-1{1/(s-a)n} = tn-1eat/(n-1)!.
Example 3: Repeated Factor
Find L-1{(2s+3)/(s+1)²}.
Step 1: (2s+3)/(s+1)² = A/(s+1) + B/(s+1)².
Step 2: 2s+3 = A(s+1) + B. Expand: As + (A+B) = 2s + 3. So A=2, B=1.
Answer: L-1 = 2e-t + te-t.
Partial Fractions: Irreducible Quadratic Factors
Irreducible Quadratic: For (s² + bs + c) that cannot factor over the reals, use (As+B)/(s²+bs+c). Complete the square and match to eatcos(bt) or eatsin(bt) forms.
Example 4: Completing the Square
Find L-1{(2s+5)/(s²+4s+13)}.
Step 1: Complete the square: s²+4s+13 = (s+2)² + 9.
Step 2: Rewrite numerator in terms of (s+2): 2s+5 = 2(s+2) + 1.
Step 3: Split: 2(s+2)/[(s+2)²+9] + 1/[(s+2)²+9] = 2(s+2)/[(s+2)²+9] + (1/3) · 3/[(s+2)²+9].
Answer: L-1 = 2e-2tcos(3t) + (1/3)e-2tsin(3t).
Mixed Factor Types
Example 5: Linear and Quadratic Factors
Find L-1{5/[s(s²+4)]}.
Step 1: 5/[s(s²+4)] = A/s + (Bs+C)/(s²+4).
Step 2: 5 = A(s²+4) + (Bs+C)s. Set s=0: 5=4A, A=5/4.
Expand: (A+B)s² + Cs + 4A = 5. So A+B=0 gives B=-5/4, and C=0.
Step 3: F(s) = (5/4)/s - (5/4)s/(s²+4).
Answer: L-1 = (5/4) - (5/4)cos(2t).
Completing the Square: Summary Strategy
When the denominator is s² + bs + c:
- Rewrite as (s + b/2)² + (c - b²/4). Let ω² = c - b²/4.
- Rewrite the numerator in terms of (s + b/2).
- Match to (s+p)/[(s+p)²+ω²] → e-ptcos(ωt) and ω/[(s+p)²+ω²] → e-ptsin(ωt).
Check Your Understanding
1. Find L-1{3/(s+5)}.
2. Find L-1{(s+3)/[(s+1)(s+4)]}.
3. Find L-1{s/(s²+6s+10)}.
Key Takeaways
- The inverse Laplace transform recovers f(t) from F(s) using the table in reverse.
- Partial fractions decompose rational F(s) into simpler terms matching the table.
- Three types of factors: distinct linear → A/(s-a), repeated → terms up to Ak/(s-a)k, quadratic → (As+B)/(s²+bs+c).
- Completing the square converts s²+bs+c to (s+p)²+ω², yielding e-ptcos/sin forms.
- These techniques are the essential toolkit for solving IVPs via Laplace transforms.