Lesson 3: Solving IVPs with Laplace Transforms
Estimated time: 40-50 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Apply the Laplace transform to both sides of a differential equation
- Use the derivative property to incorporate initial conditions automatically
- Solve the resulting algebraic equation for Y(s)
- Invert Y(s) using partial fractions to obtain y(t)
- Solve first-order and second-order IVPs using the full Laplace procedure
The Laplace Transform Procedure
Procedure for Solving IVPs:
1. Take the Laplace transform of both sides of the ODE.
2. Use L{y'} = sY - y(0) and L{y''} = s²Y - sy(0) - y'(0) to substitute.
3. Solve the algebraic equation for Y(s).
4. Find y(t) = L-1{Y(s)} using partial fractions and the table.
The beauty: initial conditions are built into step 2 automatically. No need to find the general solution first and then apply ICs.
First-Order Examples
Example 1: First-Order IVP
Solve y' + 3y = 6, y(0) = 2.
Step 1: Transform: sY - y(0) + 3Y = 6/s. So sY - 2 + 3Y = 6/s.
Step 2: (s+3)Y = 6/s + 2 = (6 + 2s)/s.
Step 3: Y = (2s+6)/[s(s+3)].
Step 4: Partial fractions: (2s+6)/[s(s+3)] = A/s + B/(s+3). 2s+6 = A(s+3) + Bs. s=0: 6=3A, A=2. s=-3: 0=-3B, B=0.
Answer: Y = 2/s, so y(t) = 2. Let us verify: y'=0, y'+3y = 0+6 = 6. Check: y(0) = 2. Verified!
(Actually, let me recompute. 2s+6 = 2(s+3), so Y = 2(s+3)/[s(s+3)] = 2/s. y = 2. This is correct.)
Example 2: First-Order IVP with Exponential
Solve y' - 2y = e3t, y(0) = 1.
Step 1: sY - 1 - 2Y = 1/(s-3).
Step 2: (s-2)Y = 1/(s-3) + 1 = (s-3+1)/(s-3) = (s-2)/(s-3).
Step 3: Y = 1/(s-3).
Answer: y(t) = e3t.
Second-Order Examples
Example 3: Second-Order IVP
Solve y'' + 4y = 0, y(0) = 3, y'(0) = -2.
Step 1: s²Y - sy(0) - y'(0) + 4Y = 0. So s²Y - 3s + 2 + 4Y = 0.
Step 2: (s²+4)Y = 3s - 2.
Step 3: Y = (3s-2)/(s²+4) = 3s/(s²+4) - 2/(s²+4) = 3 · s/(s²+4) - 1 · 2/(s²+4).
Answer: y(t) = 3cos(2t) - sin(2t).
Example 4: Nonhomogeneous Second-Order
Solve y'' + 5y' + 6y = 2e-t, y(0) = 0, y'(0) = 0.
Step 1: s²Y + 5sY + 6Y = 2/(s+1).
Step 2: Y(s²+5s+6) = 2/(s+1). Y = 2/[(s+1)(s+2)(s+3)].
Step 3: Partial fractions: A/(s+1) + B/(s+2) + C/(s+3).
2 = A(s+2)(s+3) + B(s+1)(s+3) + C(s+1)(s+2).
s=-1: 2 = A(1)(2) = 2A, A=1. s=-2: 2 = B(-1)(1) = -B, B=-2. s=-3: 2 = C(-2)(-1) = 2C, C=1.
Answer: y(t) = e-t - 2e-2t + e-3t.
Check Your Understanding
1. What is the first step when solving an IVP using Laplace transforms?
2. Solve y' + y = 0, y(0) = 5 using Laplace transforms.
3. Why is the Laplace method advantageous over traditional methods for IVPs?
Key Takeaways
- Transform both sides: derivatives become polynomials in s times Y(s), minus initial values.
- Solve algebraically for Y(s) -- no differential equations in the s-domain.
- Invert using partial fractions and the table to find y(t).
- Initial conditions are built in automatically, giving the particular solution directly.
- The method works for any order, any constant-coefficient linear ODE.