Solving an IVP by Transform, End to End

Four guides, one machine

You now hold every part of the engine. Guide 1 explained what the Laplace transform does: it carries a function of time t into a new function of a variable s, trading the world of calculus for the world of algebra. Guide 2 gave you a working table so you never have to compute an integral by hand. Guide 3 showed the single most important entry — the transform of a derivative — which quietly pulls the initial conditions inside. Guide 4 taught the return trip: the inverse transform, unlocked by partial fractions. This guide bolts the parts together and turns the key.

The reason this is worth a whole method is the deep trick at its heart: the transfer of an IVP to algebra. A differential equation asks you to undo derivatives, which is hard. The transform replaces every derivative with multiplication by s, so the whole equation becomes a polynomial relation in s — something you solved in school. The hard part of calculus has been swapped for the easy part of algebra. That swap is the entire payoff.

The five-step algorithm, named once

Every Laplace solution of a linear constant-coefficient initial value problem follows the same shape. It is worth memorising as a single rhythm — transform, substitute, solve, decompose, invert — because once it is a reflex you stop thinking about the machinery and start thinking about the problem. We call the whole routine the master algorithm.

1. Transform the whole equation. Apply the transform to both sides; by linearity it goes term by term.
2. Substitute the initial conditions. Use the derivative rule, where y(0) and y'(0) appear automatically — there is no separate step to impose them.
3. Solve for Y(s) algebraically. The equation is now linear in the single unknown Y(s); just isolate it.
4. Decompose with partial fractions. Split the rational Y(s) into the simple pieces your table recognises.
5. Invert term by term. Read each piece backwards through the table to recover y(t).

A full example, watched step by step

Take the forced spring you met back in the oscillations rung: y'' + 4y = 0, with y(0) = 3 and y'(0) = 2. Step 1, transform both sides. The derivative rule from guide 3 says the transform of y'' is s^2 Y(s) - s y(0) - y'(0). The transform of 4y is just 4 Y(s). So the equation becomes s^2 Y(s) - s y(0) - y'(0) + 4 Y(s) = 0.

Step 2, substitute the data. This is the moment that makes the transform special: y(0) = 3 and y'(0) = 2 are already sitting in the equation, so we just drop them in to get s^2 Y(s) - 3s - 2 + 4 Y(s) = 0. The initial conditions are built in — they entered the algebra the instant we transformed the derivatives, not at the end as an afterthought.

y'' + 4y = 0,   y(0) = 3,  y'(0) = 2

[1] transform :  s^2 Y - s y(0) - y'(0)  +  4 Y  =  0
[2] put data  :  s^2 Y - 3s - 2  +  4 Y    =  0
[3] solve Y   :  (s^2 + 4) Y = 3s + 2
                 Y(s) = (3s + 2) / (s^2 + 4)
[4] split     :  Y(s) = 3 * s/(s^2+4)  +  1 * 2/(s^2+4)
[5] invert    :  y(t) = 3 cos(2t) + 1 sin(2t)

Solving for Y(s), then reading it backwards

Step 3 is pure algebra. Gather the Y(s) terms: (s^2 + 4) Y(s) = 3s + 2, so Y(s) = (3s + 2) / (s^2 + 4). The factor (s^2 + 4) is no accident — it is exactly the characteristic polynomial s^2 + 4 of the original equation, now appearing as a denominator. The transform has quietly handed you the same structure you would have found by the characteristic-root method, only it arrived as ordinary fraction algebra.

Steps 4 and 5 are the return trip. Here the denominator is already a single irreducible piece, so the 'partial fraction' work is just splitting the numerator: Y(s) = 3 * s/(s^2 + 4) + 2/(s^2 + 4). Now read each term backwards through the table. The pair s/(s^2 + 4) is the transform of cos(2t), and 2/(s^2 + 4) is the transform of sin(2t). Therefore y(t) = 3 cos(2t) + sin(2t). Done — no constants left to chase.

What the method is, and what it is not

Be honest about the boundaries. The master algorithm in this clean form is built for linear equations with constant coefficients, where transforming really does turn derivatives into multiplication by s. Variable coefficients break that promise — the transform of t·y(t) involves differentiating in s, and the tidy polynomial picture dissolves. And the method is not magic for nonlinear equations: the transform of y^2 is not the square of Y(s), so the whole approach simply does not apply there.

Even within its home turf there is fine print. Inverting Y(s) cleanly assumes the denominator factors into pieces your table covers; a partial-fraction expansion with repeated or complex roots still works, but it asks for more care, often leaning on the shifting theorem to handle the e^(at) factors that complex or shifted roots produce. None of this is a flaw — it is the same trade you accept everywhere in ODEs: a tidy method covers the lucky linear cases, and the wild rest needs other tools.

What you should carry away is the shape of the idea, not just this one example. The transform takes you out of the domain where the problem is hard, lets you do something easy there, and brings you back. That round-trip pattern — hard problem, easy detour, return — is one of the most powerful moves in all of applied mathematics, and the next rung shows it doing something the classical methods could barely touch: forcing that switches on, instantaneous kicks, and the convolution that links input to output.