The Laplace transform, explained
The Laplace transform turns a differential equation in time into an algebraic equation in a new variable, s. You solve the algebra, then transform back. It is how linear ODEs with initial conditions, circuits and control systems are solved.
What it does
The transform multiplies a function by e^(−st) and integrates over all positive time. The property that makes it useful is what it does to derivatives: a derivative becomes multiplication by s, and the initial conditions drop in automatically.
Solving an ODE in three steps
Transform both sides of the equation, using the derivative rules and the table. Solve the resulting algebraic equation for Y(s). Then turn Y(s) back into y(t), usually by splitting it into partial fractions that each appear in the table.
Inverting with partial fractions
Most second-order problems end in a fraction (As + B)/(s² + ps + q). The roots of the denominator decide the answer. Two real roots give two exponentials. A repeated root gives a term in t·e^(rt). Complex roots give a decaying sine and cosine, found by completing the square. The calculator below handles all three cases.
Worked example
Solve y″ + 3y′ + 2y = 0 with y(0) = 1 and y′(0) = 0. Transforming gives s²Y − s + 3(sY − 1) + 2Y = 0, so Y(s) = (s + 3)/(s² + 3s + 2).
The denominator is (s + 1)(s + 2), and partial fractions give Y = 2/(s + 1) − 1/(s + 2). So y(t) = 2e^(−t) − e^(−2t). Check: y(0) = 2 − 1 = 1 and y′(0) = −2 + 2 = 0. These are the calculator’s default inputs.
Inverse Laplace calculator
F(s) = (A·s + B) / (s² + p·s + q)
Common mistakes
- Dropping the initial conditions when transforming a derivative.
- Getting the shift backwards: multiplying by e^(at) in time replaces s with s − a, not s + a.
- Starting partial fractions when the numerator’s degree is not lower than the denominator’s. Divide the polynomials first.