Solving ODEs with Laplace Transforms

Solving a differential equation the classical way is a two-part chore: first hunt down the general solution (the homogeneous part plus a particular part), then wrestle the arbitrary constants until they match the initial conditions. The Laplace transform offers a completely different deal. It hands you a machine that turns the whole initial-value problem — differential equation and starting data together — into an ordinary algebra problem. You solve the algebra, then transform back. No two-part bookkeeping, no fitting constants at the end.

This is why engineers reach for Laplace transforms first: they swallow a discontinuous input (a switch flicked on at t = 3, a sudden hammer-blow) without complaint, and the initial conditions are baked straight into the equations from the very first line.

The one rule that makes it work

Everything hinges on how the Laplace transform treats a derivative. Integration by parts turns a derivative in t into multiplication by s — and, crucially, drops the starting value out as a leftover term. Writing Y(s) = \mathcal{L}\{y(t)\}:

\mathcal{L}\{y'\} = sY(s) - y(0)
\mathcal{L}\{y''\} = s^2 Y(s) - s\,y(0) - y'(0)

Read those again and notice the magic: the initial values y(0) and y'(0) ride along inside the transform. The moment you transform the equation, the starting data is already in it. There is no separate step where you "apply the initial conditions" — they were never left out.

The method in three moves

Every Laplace solution of an initial-value problem follows the same pipeline. It never changes:

Transform. Take \mathcal{L}\{\,\cdot\,\} of the whole equation, using the derivative rules so the initial conditions drop in automatically.
Solve for Y(s). What's left is a plain algebraic equation in Y(s) — no calculus. Rearrange to get Y(s) by itself.
Invert. Take the inverse transform y(t) = \mathcal{L}^{-1}\{Y(s)\} — usually by breaking Y(s) into partial fractions that each match a line of the transform table.

The differential equation goes in one end; the solution y(t) — already satisfying its initial conditions — comes out the other. Here is that pipeline drawn out:

Worked example 1 — a first-order equation

Solve y' + 2y = 0 with y(0) = 3.

Transform. Apply \mathcal{L} term by term. Using \mathcal{L}\{y'\} = sY - y(0) = sY - 3:

\bigl(sY - 3\bigr) + 2Y = 0.

Solve for Y(s). Gather the Y terms — this is now pure algebra:

(s + 2)\,Y = 3 \;\Longrightarrow\; Y(s) = \frac{3}{s + 2}.

Invert. The table entry \mathcal{L}\{e^{at}\} = \dfrac{1}{s - a} matches with a = -2, so

y(t) = 3\,e^{-2t}.

Check it: y(0) = 3e^{0} = 3 ✓, and y' + 2y = -6e^{-2t} + 6e^{-2t} = 0 ✓. Notice we never introduced a constant of integration and never "applied the initial condition" as a separate step — the 3 was carried in by the derivative rule from the first line.

Worked example 2 — a second-order equation

Solve y'' + y = 0 with y(0) = 0 and y'(0) = 1.

Transform. Now use \mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0) = s^2 Y - s(0) - 1 = s^2 Y - 1:

\bigl(s^2 Y - 1\bigr) + Y = 0.

Solve for Y(s).

(s^2 + 1)\,Y = 1 \;\Longrightarrow\; Y(s) = \frac{1}{s^2 + 1}.

Invert. The table entry \mathcal{L}\{\sin t\} = \dfrac{1}{s^2 + 1} matches exactly, so

y(t) = \sin t.

A denominator of s^2 + 1 — an irreducible quadratic — is the transform's fingerprint for an oscillation. Roots of the denominator (the "poles") on the imaginary axis mean sines and cosines; a real root like s = -2 above meant a decaying exponential. The recovered solutions look like this:

Real problems have a right-hand side: a driving force. Take y'' - y = e^{t} with y(0) = 0, y'(0) = 0. Transforming (with \mathcal{L}\{e^{t}\} = \tfrac{1}{s-1}) gives (s^2 - 1)Y = \tfrac{1}{s-1}, so

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

That matches no single table entry — so we split it with \dfrac{1}{(s-1)^2(s+1)} = \dfrac{A}{s-1} + \dfrac{B}{(s-1)^2} + \dfrac{C}{s+1}. Working the partial fractions out gives A = -\tfrac14, B = \tfrac12, C = \tfrac14, and each piece is now a table entry:

y(t) = -\tfrac14 e^{t} + \tfrac12 t\,e^{t} + \tfrac14 e^{-t}.

Notice how the repeated factor (s-1)^2 produced the t\,e^{t} term — the transform automatically supplied the "resonance" piece that the classical method has to guess at. Partial fractions are the engine that turns any messy Y(s) into recognisable table entries.

Three reasons, all of them practical:

Initial conditions are free. They ride into the equations through the derivative rules, so there is no separate constant-fitting stage that goes wrong.
Discontinuous inputs are no problem. A switch turned on at t = 3, an impulse, a square-wave drive — the transform of a step or a spike is a tidy algebraic expression, so a forcing term that would terrify the classical method is just another piece of the algebra.
It turns calculus into algebra. A whole differential equation becomes a rational function of s. Circuit designers and control engineers work almost entirely in this "s-domain," where poles and zeros of Y(s) read off the stability and response of a system at a glance.

Two slips catch almost everyone learning this method:

No constants of integration. Because \mathcal{L}\{y'\} = sY - y(0) already carries the initial value, the starting data is inside the transformed equation. Do not also tack on a +\,C or try to "apply the initial conditions" at the end — that would double-count them. With Laplace, the answer you invert is already the specific solution.
You must invert. Solving for Y(s) is only two-thirds of the job — Y(s) lives in the s-domain, not the t-domain. You have to compute \mathcal{L}^{-1}\{Y(s)\} to get an actual function of t. This is where most errors happen: to match the table you nearly always split Y(s) into partial fractions first, and a wrong-shaped decomposition (forgetting the second term of a repeated factor, or a linear numerator over a quadratic) gives a wrong y(t).