Solving initial value problems with Laplace transforms
Transform the equation, solve for the unknown transform algebraically, then transform back. Initial conditions are built in from the start.
The three-step method
1. Take the Laplace transform of both sides, using the derivative formulas with the initial conditions. 2. Solve the resulting algebraic equation for \(Y(s)\). 3. Find \(y(t) = \mathcal{L}^{-1}\{Y\}\), usually by partial fractions and the table. No separate step is needed to find constants.
Worked example
Solve \(y' + 2y = e^{-t}\), \(y(0) = 3\). Transforming: \(sY - 3 + 2Y = \frac{1}{s + 1}\), so \(Y = \frac{3}{s + 2} + \frac{1}{(s + 1)(s + 2)}\). Partial fractions: \(\frac{1}{(s+1)(s+2)} = \frac{1}{s+1} - \frac{1}{s+2}\). So \(Y = \frac{2}{s+2} + \frac{1}{s+1}\) and \(y = 2e^{-2t} + e^{-t}\).
A second-order example
Solve \(y'' + 4y = 0\), \(y(0) = 1\), \(y'(0) = 2\). Transforming: \(s^2Y - s - 2 + 4Y = 0\), so \(Y = \frac{s + 2}{s^2 + 4} = \frac{s}{s^2 + 4} + \frac{2}{s^2 + 4}\). Inverting: \(y = \cos 2t + \sin 2t\).
Inverse transforms and partial fractions
Most \(Y(s)\) are rational functions. Split them into simple fractions matching table entries: \(\frac{A}{s - a}\) gives \(Ae^{at}\); \(\frac{A}{(s - a)^2}\) gives \(Ate^{at}\); quadratic factors give sines and cosines, after completing the square if needed.
Completing the square
For \(Y = \frac{s + 3}{s^2 + 2s + 5}\), complete the square: \(s^2 + 2s + 5 = (s + 1)^2 + 4\). Write \(s + 3 = (s + 1) + 2\). Then \(Y = \frac{s+1}{(s+1)^2 + 4} + \frac{2}{(s+1)^2 + 4}\), giving \(y = e^{-t}(\cos 2t + \sin 2t)\).
Transfer functions
For a system \(ay'' + by' + cy = g(t)\) starting at rest, \(Y(s) = \frac{1}{as^2 + bs + c}G(s)\). The factor \(H(s) = \frac{1}{as^2 + bs + c}\) is the transfer function. Engineers analyse stability and response by studying the poles of \(H(s)\), the roots of its denominator.
H(s) = \frac{1}{as^2 + bs + c}Why engineers love Laplace
Laplace transforms handle initial conditions automatically, discontinuous inputs easily, and system analysis through transfer functions. Circuit analysis replaces capacitors and inductors with \(s\)-domain impedances, so whole circuits are solved with algebra.
Common mistakes
Dropping initial-condition terms. Errors in partial fractions. Forgetting to complete the square before inverting quadratic denominators. Inverting \(\frac{1}{s^2 + 4}\) as \(\sin 2t\) instead of \(\frac{1}{2}\sin 2t\).
Key takeaways
Transform, solve algebraically for \(Y(s)\), and invert with partial fractions and the table. Initial conditions enter at the first step. Complete the square for quadratic denominators. Transfer functions \(H(s)\) describe how systems respond, and their poles determine stability.
Flashcard
Transform, solve for Y, invert
Check your understanding
Taking the Laplace transform of y' - 3y = 0 with y(0) = 2 gives: