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Solving Initial-Value Problems

Transform the whole differential equation at once, watch every derivative collapse into a power of s while the initial conditions slot themselves in for free, solve the resulting algebra, and invert by partial fractions. One clean recipe replaces every guessing trick you learned for constant-coefficient ODEs.

Transforming a whole differential equation at once

In the previous guide you earned the one rule that makes the whole machine run: the transform of a derivative turns differentiation in time into multiplication by s in the s-domain, dragging the starting value along with it. Written out, L{y'(t)} = s Y(s) - y(0), and one more application gives L{y''(t)} = s^2 Y(s) - s y(0) - y'(0). Each prime on y becomes a power of s on Y, and each application sheds one piece of initial data into the open. This guide does one thing: it points that rule at a real initial-value problem and walks the answer out the other side.

Take a concrete problem and keep it in view for the whole guide: y'' + 3 y' + 2 y = 0, with y(0) = 2 and y'(0) = -3. In your earlier rung you would attack this by writing down a characteristic equation, finding its roots, forming a general combination of exponentials, and only at the very end pinning down two constants by forcing the initial conditions. The Laplace method reverses that order in a way that feels almost unfair: the initial conditions go in at the very first step, and there are never any loose constants to chase down at the end.

Apply L to every term, using linearity to go term by term. The y'' becomes s^2 Y(s) - s y(0) - y'(0); the 3 y' becomes 3(s Y(s) - y(0)); the 2 y becomes 2 Y(s); and the right side, being zero, stays zero. Substitute the numbers y(0) = 2 and y'(0) = -3 right now, while it is easy. The transformed equation reads s^2 Y - 2 s - (-3) + 3(s Y - 2) + 2 Y = 0, which tidies to (s^2 + 3 s + 2) Y(s) - 2 s - 3 = 0. Notice there is not a single derivative left anywhere — only s and Y(s), bound together by plain arithmetic.

The initial conditions enter for free

Pause on what just happened, because it is the quiet miracle at the heart of the method. The classical approach produces a *general* solution riddled with arbitrary constants, and the initial conditions are an afterthought — a separate little linear system you solve at the very end to pin those constants down. Here, the values y(0) and y'(0) were not an afterthought at all. They walked in automatically, carried by the boundary term that fell out when integration by parts produced the derivative rule. The transform does not know how to forget them; the very act of transforming a derivative hands you the starting data, baked into the equation before you have solved anything.

One thing worth seeing clearly: the polynomial (s^2 + 3 s + 2) that multiplies Y(s) is not a coincidence — it is the characteristic polynomial of the ODE wearing a different hat. The old method asked you to set that polynomial to zero and find its roots; the new method keeps it whole and divides by it. The roots, s = -1 and s = -2, are still where the action is, but now they show up as the *poles* of Y(s) — the places where the denominator vanishes — and each pole will hand you exactly one exponential in the final answer. Same physics, same numbers, read off a different page.

Solving the algebra, then inverting by partial fractions

Now the easy middle step. From (s^2 + 3 s + 2) Y(s) - 2 s - 3 = 0, just isolate Y(s) the way you isolated x in middle school: Y(s) = (2 s + 3)/(s^2 + 3 s + 2). That is the entire 's-domain solution'. There is no calculus in it and nothing left to integrate — the whole differential equation has dissolved into a single ratio of polynomials. The only remaining task is to climb back across the bridge: find the function of t whose transform is exactly this F(s). That reverse trip is the inverse Laplace transform, and for a ratio of polynomials it is almost always done by one technique.

That technique is partial-fraction decomposition — the same algebra you first met in Volume I to tame a rational integrand, now repurposed as the workhorse of inversion. Factor the denominator: s^2 + 3 s + 2 = (s + 1)(s + 2). The point of partial fractions is that a clumsy single fraction over a product of factors can be split into a sum of simple pieces, one per factor: (2 s + 3)/((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2). We split it precisely because each simple piece A/(s - a) is a *recognizable table entry* — it is the transform of an exponential — whereas the lumped fraction matches nothing in the table.

Y(s) = (2 s + 3) / ((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2)

Clear denominators:   2 s + 3 = A (s + 2) + B (s + 1)

Cover-up at s = -1:   2(-1) + 3 = A(-1 + 2)  ->  1 = A      ->  A = 1
Cover-up at s = -2:   2(-2) + 3 = B(-2 + 1)  -> -1 = -B     ->  B = 1

  Y(s) = 1/(s + 1) + 1/(s + 2)

Read each piece backward off the table  ( 1/(s - a)  <->  e^{a t} ):

  1/(s + 1) = 1/(s - (-1))  ->  e^{-t}
  1/(s + 2) = 1/(s - (-2))  ->  e^{-2 t}

  y(t) = e^{-t} + e^{-2 t}

Check:  y(0) = 1 + 1 = 2  (matches),  y'(0) = -1 - 2 = -3  (matches).
The full inversion: split into table-sized pieces with the cover-up shortcut, then read each piece back as an exponential. The two checks at the bottom confirm the initial conditions are already satisfied — we never solved for any extra constants.

Stand back and admire the finished route. The coefficients A and B came out by the cover-up method: to find the weight sitting over the factor (s + 1), set s = -1 (the value that kills that factor) in everything except that factor itself, and read the number off. Each clean piece 1/(s - a) is then recognized straight from your table of pairs as e^{a t}, and linearity lets you add the pieces back up. The final answer y(t) = e^{-t} + e^{-2 t} already obeys both initial conditions — you can check, but you do not have to, because they were sealed in at step one.

The recipe, and what each ingredient is for

The worked example was deliberately plain so the skeleton would show through. Here is that skeleton, the universal recipe for solving an IVP by the Laplace transform. It works without modification for any linear constant-coefficient equation of any order — second, fourth, whatever — and for any forcing term f(t) on the right whose transform you can write down.

  1. Transform every term. Apply L to both sides and go term by term using linearity. Each derivative becomes a power of s times Y(s); the derivative rule drops the initial values y(0), y'(0), ... into the equation automatically. Substitute the known numbers immediately.
  2. Solve for Y(s). You now have one ordinary algebraic equation with no derivatives left. Divide through by the characteristic polynomial to get Y(s) alone — a single ratio of polynomials in s, the s-domain answer.
  3. Decompose. Factor the denominator and split Y(s) by partial fractions into a sum of simple pieces, one per factor, sized to match entries in the transform table.
  4. Invert and assemble. Read each piece backward off the table into a function of t — exponentials, sines, cosines — and add them up. The sum y(t) is the unique solution, and its initial conditions are already correct.

Two refinements to keep in your pocket, because real problems will throw them at you. When the denominator has a *repeated* factor like (s + 1)^2, the decomposition needs two terms over it, A/(s + 1) + B/(s + 1)^2; the second inverts not to a bare exponential but to t e^{-t}, which is exactly how a repeated root produces a t-times-exponential in the classical damped-oscillator story. And when the denominator has an *irreducible quadratic* like s^2 + omega^2 that will not factor over the reals, you do not force it apart — you complete the square and match it to the sine and cosine pairs, giving oscillation in t. Repeated roots make resonance-like growth; complex roots make ringing.

What is true, what is assumed, and where it stops

It is worth being honest about why this works and where it stops, so you trust it for the right reasons. The method's whole leverage is that the derivative rule is *linear* and has *constant coefficients* in mind: it cleanly converts d/dt into multiplication by s only because the coefficients out front do not depend on t. The instant a coefficient varies with t — an equation like t y'' + y' = 0 — transforming a product like t y'' no longer gives a clean power of s but a *derivative in s*, and the tidy algebra breaks down. For genuinely nonlinear equations, the transform of y^2 is not the square of the transform, and the bridge collapses entirely. This is a precision instrument for linear, constant-coefficient problems.

Finally, see what you have actually gained over the old way, because it is more than a shortcut. The classical route splits a forced problem into a homogeneous part plus a particular part, hunts the particular solution by undetermined coefficients, and then reconciles everything with the initial conditions — three separate skills. The Laplace recipe fuses all of that into one straight-line computation: the forcing f(t) and the initial data ride into the same transformed equation together, and a single Y(s) already carries both the natural response and the forced response, correctly weighted. The next guides extend this same machine to forcing that switches on and off or strikes as an instantaneous impulse — the cases where the classical method grows truly painful and the transform stays effortless.