Solving Initial Value Problems with Laplace Transforms

The Complete Pipeline: Transform → Algebra → Inverse

Learning Objectives
  • Apply derivative transform formulas to convert ODEs into algebraic equations
  • Set up and solve algebraic equations in the s-domain
  • Incorporate initial conditions automatically without separate solving
  • Extract transfer functions from ODE systems

This is where everything comes together. The Laplace transform method turns solving a differential equation into a 3-step algebraic process:

  1. Transform both sides of the ODE using derivative rules
  2. Solve algebraically for $Y(s)$ in the s-domain
  3. Inverse transform to recover $y(t)$

Initial conditions are baked right into step 1 — no need for a separate step to find constants like $C_1$ and $C_2$. This is a powerful advantage over classical methods.

"Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."

— Albert Einstein (1879–1955)

يَرْفَعِ اللَّهُ الَّذِينَ آمَنُوا مِنكُمْ وَالَّذِينَ أُوتُوا الْعِلْمَ دَرَجَاتٍ

"Allah will raise those who have believed among you and those who were given knowledge, by degrees."

— Quran, Surah Al-Mujadila 58:11

Theory: Key Formulas

The Transform of Derivatives

First Derivative: $$\mathcal{L}\{y'\} = sY(s) - y(0)$$ Second Derivative: $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$
Engineering Insight: Notice how initial conditions appear naturally! The term $sy(0)$ represents the "energy" stored in the system at $t=0$. For a capacitor, this is the initial voltage; for a spring, the initial displacement. The Laplace method handles initial conditions automatically — no need to solve for $C_1$, $C_2$ separately.

The Transfer Function

Definition: For the ODE $ay'' + by' + cy = f(t)$ with zero initial conditions, transforming gives: $$Y(s) = \frac{1}{as^2 + bs + c} \cdot F(s) = G(s) \cdot F(s)$$ where $G(s) = \dfrac{1}{as^2 + bs + c}$ is the transfer function.

The transfer function $G(s)$ is the system's "fingerprint" — it tells you how ANY input $f(t)$ will be processed. In the s-domain: Output = Transfer Function × Input.

Complete Solution Structure

For $ay'' + by' + cy = f(t)$, $y(0) = y_0$, $y'(0) = y_1$: $$(as^2 + bs + c)Y(s) = F(s) + (as + b)y_0 + ay_1$$ $$Y(s) = \underbrace{\frac{(as+b)y_0 + ay_1}{as^2+bs+c}}_{\text{Natural response}} + \underbrace{\frac{F(s)}{as^2+bs+c}}_{\text{Forced response}}$$
Connection to Classical Methods: This is the same $y_c + y_p$ split from Chapter 2! The natural response comes from initial conditions, and the forced response comes from the input forcing function. Laplace makes this decomposition crystal clear.

Step-by-Step Solution Method

  1. Take the Laplace transform of both sides of the ODE
  2. Substitute the derivative formulas: $\mathcal{L}\{y'\} = sY - y(0)$, $\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$
  3. Substitute the known initial conditions $y(0)$ and $y'(0)$
  4. Solve algebraically for $Y(s)$
  5. Use partial fractions to decompose $Y(s)$ into simple terms
  6. Apply the inverse Laplace transform to find $y(t)$
Key Tip: The success of this method depends on:
  • Correctly applying the derivative rules with initial conditions
  • Skillful partial fraction decomposition of $Y(s)$
  • Familiarity with the standard Laplace transform pairs

Worked Examples

Step 1: Transform both sides

$$\mathcal{L}\{y' + 2y\} = \mathcal{L}\{6\}$$ $$(sY - 1) + 2Y = \frac{6}{s}$$

Step 2: Solve for $Y(s)$

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

Step 3: Partial fractions

$$\frac{6}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2}$$ $$6 = A(s+2) + Bs \quad \Rightarrow \quad A = 3, \, B = -3$$ $$Y = \frac{3}{s} - \frac{3}{s+2} + \frac{1}{s+2} = \frac{3}{s} - \frac{2}{s+2}$$

Step 4: Inverse transform

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

✓ Verification:

At $t=0$: $y(0) = 3 - 2 = 1$ ✓. As $t \to \infty$: $y \to 3$ (steady state). Check ODE: $y' = 4e^{-2t}$, so $y' + 2y = 4e^{-2t} + 2(3 - 2e^{-2t}) = 6$ ✓

Step 1: Transform both sides

$$\mathcal{L}\{y'' - 3y' + 2y\} = 0$$ $$(s^2Y - s + 0) - 3(sY - 1) + 2Y = 0$$

Step 2: Solve for $Y(s)$

$$(s^2 - 3s + 2)Y = s - 3$$ $$Y = \frac{s-3}{(s-1)(s-2)}$$

Step 3: Partial fractions using Heaviside cover-up

$$A = \frac{1-3}{1-2} = 2, \quad B = \frac{2-3}{2-1} = -1$$ $$Y = \frac{2}{s-1} - \frac{1}{s-2}$$

Step 4: Inverse transform

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

Step 1: Transform

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

Step 2: Partial fractions

$$\frac{1}{(s^2+1)(s^2+4)} = \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+4}$$

Multiplying and matching coefficients: $A = 0$, $B = 1/3$, $C = 0$, $D = -1/3$

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

Step 3: Inverse transform

$$y(t) = \frac{1}{3}\sin(t) - \frac{1}{6}\sin(2t)$$

Interpretation:

The solution is a superposition of two sine waves at different frequencies: the forcing frequency (1 rad/s) and the natural frequency (2 rad/s). This is beat phenomena.

Step 1: Transform

$$s^2Y - 2s + 1 + 2(sY - 2) + 5Y = 0$$ $$(s^2 + 2s + 5)Y = 2s + 3$$

Step 2: Notice the quadratic structure

$$s^2 + 2s + 5 = (s+1)^2 + 4$$

Step 3: Rewrite numerator in terms of $(s+1)$

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

Step 4: Inverse transform using shift theorem

$$y(t) = 2e^{-t}\cos(2t) + \frac{1}{2}e^{-t}\sin(2t)$$

Interpretation:

Underdamped oscillation ($\zeta = 0.316 < 1$) with exponential decay envelope $e^{-t}$ and oscillation frequency 2 rad/s.

Step 1: Transform

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

Step 2: Partial fractions for the second term

$$\frac{s}{(s^2+1)(s^2+4)} = \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+4}$$

Matching: $A = 1/3$, $B = 0$, $C = -1/3$, $D = 0$

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

Step 3: Inverse transform

$$y(t) = \frac{4}{3}\cos(t) - \frac{1}{3}\cos(2t)$$

Key Point:

Even though the forcing frequency (2 rad/s) is different from the natural frequency (1 rad/s), there's NO resonance. The system oscillates at both frequencies simultaneously.

Step 1: Transform

$$s^2Y - 0 - 1 - Y = \frac{1}{s-2}$$ $$Y(s^2 - 1) = 1 + \frac{1}{s-2}$$

Step 2: Solve

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

Step 3: Combine and simplify

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

Step 4: Partial fractions

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

Step 1: Transform

$$s^2Y - 2s + 3 + 6(sY - 2) + 9Y = 0$$ $$(s^2 + 6s + 9)Y = 2s + 9$$

Step 2: Factor the denominator

$$Y = \frac{2s + 9}{(s+3)^2}$$

Step 3: Partial fractions with repeated root

$$\frac{2s + 9}{(s+3)^2} = \frac{A}{s+3} + \frac{B}{(s+3)^2}$$ $$2s + 9 = A(s+3) + B$$ $$\text{At } s = -3: \, B = 3. \quad \text{Matching } s: A = 2$$ $$Y = \frac{2}{s+3} + \frac{3}{(s+3)^2}$$

Step 4: Inverse transform

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

Critical damping ($\zeta = 1$):

The repeated root at $s = -3$ indicates critical damping. The solution has a polynomial factor $t$ in addition to the exponential.

Problem: An RLC circuit with $R = 4\Omega$, $L = 1\text{H}$, $C = 1/4\text{F}$ is subjected to a constant voltage. The equation for charge is:

$$y'' + 4y' + 4y = 10, \quad y(0) = 0, \quad y'(0) = 0$$

Step 1: Transform

$$s^2Y + 4sY + 4Y = \frac{10}{s}$$ $$(s+2)^2Y = \frac{10}{s}$$ $$Y = \frac{10}{s(s+2)^2}$$

Step 2: Partial fractions

$$\frac{10}{s(s+2)^2} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{(s+2)^2}$$ $$10 = A(s+2)^2 + Bs(s+2) + Cs$$

At $s=0$: $10 = 4A \Rightarrow A = 5/2$

At $s=-2$: $10 = -2C \Rightarrow C = -5$

Matching $s^2$: $0 = A + B \Rightarrow B = -5/2$

$$Y = \frac{5/2}{s} - \frac{5/2}{s+2} - \frac{5}{(s+2)^2}$$

Step 3: Inverse transform

$$y(t) = \frac{5}{2} - \frac{5}{2}e^{-2t} - 5te^{-2t}$$

Physical Interpretation:

  • Steady-state charge: $y_{\text{ss}} = 5/2 = 2.5$ coulombs (from $\lim_{t \to \infty} y(t)$)
  • Transient response: The $e^{-2t}$ and $te^{-2t}$ terms decay due to critical damping ($\zeta = 1$)
  • Current: $i(t) = y'(t) = 5e^{-2t} - 5e^{-2t} + 10te^{-2t} = 10te^{-2t}$, starting at 0 and decaying to 0

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: Solve $y'' + 4y = 8$ with $y(0)=0$, $y'(0)=2$ ▼

Solution

Step 1: Apply the Laplace transform to both sides. Recall: $\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$ and $\mathcal{L}\{8\} = \dfrac{8}{s}$

$$s^2Y - 0 - 2 + 4Y = \frac{8}{s}$$

Step 2: Collect terms with $Y$:

$$(s^2+4)Y = \frac{8}{s} + 2$$

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

$$(s^2+4)Y = \frac{8 + 2s}{s}$$
$$Y = \frac{8 + 2s}{s(s^2+4)} = \frac{8}{s(s^2+4)} + \frac{2s}{s(s^2+4)} = \frac{8}{s(s^2+4)} + \frac{2}{s^2+4}$$

Step 4: Use partial fractions on $\dfrac{8}{s(s^2+4)}$:

$$\frac{8}{s(s^2+4)} = \frac{A}{s} + \frac{Bs+C}{s^2+4}$$

Multiplying by $s(s^2+4)$: $8 = A(s^2+4) + (Bs+C)s$

  • At $s=0$: $8 = 4A$, so $A = 2$
  • Comparing coefficients: $B = -2$, $C = 0$
$$\frac{8}{s(s^2+4)} = \frac{2}{s} - \frac{2s}{s^2+4}$$

Step 5: Combine:

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

Step 6: Apply inverse Laplace transform:

$$\mathcal{L}^{-1}\left\{\frac{2}{s}\right\} = 2$$
$$\mathcal{L}^{-1}\left\{-\frac{2s}{s^2+4}\right\} = -2\cos(2t)$$
$$\mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\} = \sin(2t)$$

Answer: $\boxed{y(t) = 2 - 2\cos(2t) + \sin(2t)}$

Practice 2: Solve $y'' - 3y' + 2y = e^{3t}$ with $y(0)=1$, $y'(0)=0$ ▼

Solution

Step 1: Apply Laplace transform. $\mathcal{L}\{y''\} = s^2Y - s(1) - 0 = s^2Y - s$, $\mathcal{L}\{-3y'\} = -3(sY - 1) = -3sY + 3$

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

Step 2: Factor the left side:

$$(s^2-3s+2)Y = \frac{1}{s-3} + s - 3$$

Step 3: Factor $s^2-3s+2 = (s-1)(s-2)$ and simplify RHS:

$$(s-1)(s-2)Y = \frac{1 + (s-3)^2}{s-3} = \frac{1 + s^2 - 6s + 9}{s-3} = \frac{s^2 - 6s + 10}{s-3}$$

Step 4: Solve for $Y$:

$$Y = \frac{s^2-6s+10}{(s-1)(s-2)(s-3)}$$

Step 5: Use partial fractions: $\dfrac{A}{s-1} + \dfrac{B}{s-2} + \dfrac{C}{s-3}$

  • At $s=1$: $\dfrac{1-6+10}{(-1)(-2)} = \dfrac{5}{2}$, so $A = \dfrac{5}{2}$
  • At $s=2$: $\dfrac{4-12+10}{(1)(-1)} = \dfrac{2}{-1} = -2$, so $B = -2$
  • At $s=3$: $\dfrac{9-18+10}{(2)(1)} = \dfrac{1}{2}$, so $C = \dfrac{1}{2}$

Step 6: Apply inverse transform:

$$\mathcal{L}^{-1}\left\{\frac{5/2}{s-1}\right\} = \frac{5}{2}e^{t}$$
$$\mathcal{L}^{-1}\left\{\frac{-2}{s-2}\right\} = -2e^{2t}$$
$$\mathcal{L}^{-1}\left\{\frac{1/2}{s-3}\right\} = \frac{1}{2}e^{3t}$$

Answer: $\boxed{y(t) = \dfrac{5}{2}e^{t} - 2e^{2t} + \dfrac{1}{2}e^{3t}}$

Interactive ODE Solver & Transfer Function Visualizer

Adjust the parameters below to see how the transfer function poles and solution change.

Transfer Function Analysis

Transfer Function: $G(s) = 1/(s^2 + 2s + 5)$

Poles: Complex pair at $s = -1 \pm 2i$

Damping Ratio ($\zeta$): 0.316 (underdamped oscillation)

Practice Quiz

Question 1: For the IVP $y' + 3y = 6$, $y(0) = 0$, which of the following is the correct $Y(s)$?

Question 2: The transfer function of $y'' + 4y' + 3y = f(t)$ is:

Question 3: For $y'' - y = 0$, $y(0) = 1$, $y'(0) = 1$, what is $Y(s)$?

Question 4: If $Y(s) = \dfrac{2s+3}{(s+1)^2+4}$, the solution $y(t)$ involves:

Question 5: In an RLC circuit, the poles of the transfer function determine:

Question 6: For $y'' + 9y = 0$, $y(0) = 0$, $y'(0) = 3$, what is the solution $y(t)$?