Solving ODEs with Laplace Transforms: Step-by-Step

Solving an ODE with Laplace transforms follows a fixed three-step recipe: transform the entire equation (initial conditions are absorbed automatically by the derivative rules), solve the resulting algebraic equation for Y(s), and invert Y(s) back to y(t). No guessing of particular solutions is ever needed.

This uniformity is its power. The same procedure handles homogeneous equations, sinusoidal forcing, discontinuous step inputs, and Dirac-delta impulses — cases where the method of undetermined coefficients either fails or needs separate tricks. Control engineers and circuit designers use it as their default solver.

Key Concepts

• Transform of derivatives carries the data: L{y′} = sY − y(0) and L{y″} = s2Y − s·y(0) − y′(0) inject both initial values directly into the algebra.
• Algebraic equation for Y(s): a constant-coefficient ODE becomes (polynomial in s)·Y(s) = (polynomial from initial data and forcing).
• Partial fractions: split Y(s) into first-order terms A/(s−a) and quadratic terms (As+B)/((s−α)2+β2), each invertible from the table.
• Completing the square: turns s2 + bs + c with complex roots into (s−α)2 + β2, inverting to damped sines and cosines eαtcos(βt).
• Convolution: when Y(s) is a product of two transforms, y(t) is their convolution integral — the fallback when partial fractions get messy.

Formulas

• L{y′} = sY(s) − y(0); L{y″} = s2Y(s) − s·y(0) − y′(0).
• Partial fraction templates: P(s)/((s−a)(s−b)) = A/(s−a) + B/(s−b); for repeated roots include A/(s−a)2.
• Inversions: A/(s−a) → Aeat; (s−α)/((s−α)2+β2) → eαtcos(βt); β/((s−α)2+β2) → eαtsin(βt).
• Assumption: linear ODE with constant coefficients. Limit: variable coefficients generally do not transform nicely.

Worked Example

Solve y″ + 3y′ + 2y = 0 with y(0) = 1, y′(0) = 0. Transform: [s2Y − s·1 − 0] + 3[sY − 1] + 2Y = 0, giving (s2 + 3s + 2)Y = s + 3, so Y(s) = (s+3)/((s+1)(s+2)). Partial fractions: (s+3)/((s+1)(s+2)) = A/(s+1) + B/(s+2), so s + 3 = A(s+2) + B(s+1). Setting s = −1 gives 2 = A; setting s = −2 gives 1 = −B, so B = −1. Thus Y = 2/(s+1) − 1/(s+2) and y(t) = 2e−t − e−2t. Check: y(0) = 2 − 1 = 1 and y′ = −2e−t + 2e−2t gives y′(0) = 0. Substituting into the ODE: (2e−t − 4e−2t) + 3(−2e−t + 2e−2t) + 2(2e−t − e−2t) = e−t(2−6+4) + e−2t(−4+6−2) = 0. Verified.

Common Misconceptions

• Forgetting the initial-value terms. Writing L{y″} as s2Y drops −s·y(0) − y′(0) and silently solves a different problem with zero initial data.
• Laplace is only for homogeneous equations. Forcing functions transform too — including steps and impulses, which is precisely where the method shines.
• Every Y(s) inverts by table lookup. Complicated rational functions need partial fractions first; products of transforms need convolution.

Real-World Applications

• Mass-spring-damper hit by a hammer: the impulse δ(t) transforms to 1, giving the exact response in three lines.
• Switching transients in RLC circuits — the moment a switch closes is t = 0 with fresh initial conditions.
• PID controller design: the closed-loop transfer function predicts overshoot and settling time before anything is built.

To solve a constant-coefficient ODE by Laplace: transform the equation with L{y′} = sY − y(0) and L{y″} = s2Y − sy(0) − y′(0), solve algebraically for Y(s), then invert via partial fractions. Initial conditions enter automatically, and steps and impulses are handled exactly.