Solving Differential Equations with the Laplace Transform

Take the Laplace transform of both sides of the differential equation, turning the ODE into an algebraic equation in the s-domain. Solve for Y(s), the transform of the unknown function. Take the inverse Laplace transform of Y(s) to get y(t). This method converts a calculus problem into an algebra problem.

The critical move is the first step: apply the derivative transform formulas correctly. Without the initial conditions built into those formulas, the algebraic equation will be wrong and the whole solution fails.

The derivative theorem gives ℒ{y'} = sY(s) - y(0) and ℒ{y''} = s²Y(s) - s y(0) - y'(0). These formulas are the only place the initial conditions enter the Laplace method. Every higher derivative follows the pattern: ℒ{y⁽ⁿ⁾} = sⁿY(s) - sⁿ⁻¹y(0) - sⁿ⁻²y'(0) - … - y⁽ⁿ⁻¹⁾(0).

Common failure: forgetting to subtract y(0) in the first-derivative formula. If y(0) is zero the term disappears and the mistake goes undetected until the final inverse fails. Boyce & DiPrima ch. 6.2 gives the full derivation; the table below shows the two formulas you need for every IVP up to second order.

Derivative	Laplace Transform	Initial Condition
y'(t)	sY(s) - y(0)	y(0)
y''(t)	s²Y(s) - s y(0) - y'(0)	y(0), y'(0)

Solve y' + 3y = 6e^(2t), y(0) = 4.

Decompose the second term by partial fractions: 6/[(s-2)(s+3)] = A/(s-2) + B/(s+3). The Heaviside cover-up method gives A = 6/(2+3) = 6/5, B = 6/(-3-2) = -6/5. So Y(s) = 4/(s+3) + (6/5)/(s-2) - (6/5)/(s+3) = (6/5)/(s-2) + (14/5)/(s+3).

Invert using the pair 1/(s-a) ->The inverse Laplace transform table from Boyce & DiPrima Table 6.2.1 or Kreyszig Table 6.9 gives these pairs directly.

Solve y'' + 2y' + 5y = 0, y(0) = 1, y'(0) = -1.

Transform: s²Y(s) - s(1) - (-1) + 2[sY(s) - 1] + 5Y(s) = 0. Simplify: (s² + 2s + 5)Y(s) - s + 1 - 2 = 0 → Y(s) = (s + 1)/(s² + 2s + 5).

Complete the square in the denominator: s² + 2s + 5 = (s+1)² + 4. Write Y(s) = (s+1)/[(s+1)² + 4]. This matches the pair (s+a)/[(s+a)² + ω²] -> e^(-at)cos(ωt). Here a=1, ω=2: y(t) = e^(-t)cos(2t).

The complex conjugate poles at s = -1 ± 2i produce a damped sinusoid. The damping ratio ζ = 1/√5 gives an underdamped response with overshoot. Ogata's second-order system standard form ω_n²/(s²+2ζω_n s+ω_n²) shows ω_n = √5 rad/s.

Solve y'' + 4y = u(t-3) - u(t-5), y(0) = 0, y'(0) = 0, where u(t-a) is the unit step function.

Transform: s²Y(s) + 4Y(s) = e^{-3s}/s - e^{-5s}/s. Solve: Y(s) = (e^{-3s} - e^{-5s})/[s(s²+4)].

First find H(s) = 1/[s(s²+4)] and decompose: 1/[s(s²+4)] = A/s + (Bs + C)/(s²+4). Multiply: 1 = A(s²+4) + s(Bs+C). Compare coefficients: A=1/4, B=-1/4, C=0. So H(s) = (1/4)/s - (1/4)s/(s²+4).

By the time-shift theorem, the inverse of e^{-as}H(s) is h(t-a)u(t-a). Therefore y(t) = h(t-3)u(t-3) - h(t-5)u(t-5).

The response is zero until t=3, then oscillates around 0.25 until t=5, then returns to zero. The step input from 3 to 5 seconds is an on-off pulse. Missing the unit step function in the time-shifted inverse is the most common error: writing h(t-3) without u(t-3) gives a function defined for negative time.

For a system like x' = -2x + y, y' = x - 2y with x(0)=1, y(0)=0, transform each equation separately. You get two algebraic equations in X(s) and Y(s). Solve the linear system, then invert each transform. The algebra scales linearly with the number of equations; a 3×3 system with distinct real poles is still manageable by hand, but a 4×4 usually requires symbolic computation via SymPy inverse_laplace_transform or Wolfram InverseLaplaceTransform. Systems with repeated eigenvalues need the partial-fraction form for repeated poles: A/(s-p) + B/(s-p)², never A/(s-p)² alone.

Handling Messy Coefficients And Repeated Poles

When your partial fraction coefficients become messy decimals from real measured data, do not round. Keep them as fractions by rationalising. The rationalised form will match the transform pairs in Boyce & DiPrima ch. 6.2-6.4 or Kreyszig ch. 6.3-6.4 exactly. Rounding introduces a phase error that the inverse transform cannot recover.

If Y(s) has a repeated complex-conjugate pole pair, textbooks show real repeated poles and complex-conjugate poles separately but not the combination. The correct partial fraction form for (s+a)/[(s+b)²+ω²]² is (As+B)/[(s+b)²+ω²] + (Cs+D)/[(s+b)²+ω²]². Solve the four equations by equating coefficients.

Handling Time Delays

When you have a time delay e^{-sT} in the transform, factor it out. Invert the delay-free part first, then apply the time-shift theorem: the inverse is the delay-free inverse shifted by T and multiplied by u(t-T). The most common failure is writing f(t-T) without u(t-T).