Differential Equations · step 14 of 14

Laplace transforms for initial value problems

The Laplace transform turns differentiation into multiplication: L{y′} = sY(s) − y(0) and L{y″} = s²Y − sy(0) − y′(0). An initial value problem therefore becomes an algebraic equation in Y(s), which you solve and then invert. The initial conditions enter automatically, which is why the method is so well suited to discontinuous forcing.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Transform every termUse the derivative rules, which build in y(0) and y′(0) directly.
Solve algebraically for Y(s)Collect the Y terms and divide — no calculus is involved at this stage.
Decompose with partial fractionsSplit Y(s) into standard transform shapes such as 1/(s−a) and s/(s²+ω²).
Invert term by termL⁻¹{1/(s−a)} = e^(at); L⁻¹{ω/(s²+ω²)} = sin ωt; L⁻¹{n!/s^(n+1)} = tⁿ.

Worked example

Solve y′ + 3y = 0 with y(0) = 2 using Laplace transforms.

Transform: (sY − y(0)) + 3Y = 0, so sY − 2 + 3Y = 0.
Collect: Y(s + 3) = 2.
So Y = 2/(s + 3), which is already a standard form.
Invert: L⁻¹{2/(s+3)} = 2e^(−3t).

Answer. y = 2e^(−3t) — the initial condition was built into the transform, with no constant to fit afterwards.

Where marks get dropped

These are the specific errors that cost credit on laplace transforms for initial value problems questions — QED's rubric penalises each of them separately.

Forgetting the initial-condition terms in the derivative rules. L{y′} = sY − y(0), and dropping y(0) loses the entire IVP structure.
Trying to invert a rational function without decomposing it first. Partial fractions is what produces recognisable shapes.
Confusing the transforms of sin and cos: L{sin ωt} = ω/(s²+ω²) while L{cos ωt} = s/(s²+ω²) — note which numerator goes with which.

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Laplace transforms for initial value problems — frequently asked questions

When is the Laplace method best?

For discontinuous or impulsive forcing — step functions and delta functions — where classical methods require awkward piecewise work. It also handles the initial conditions automatically.

What is the shifting theorem?

L{e^(at)f(t)} = F(s − a): multiplying by an exponential in t shifts the transform in s. It is what lets you invert terms like 1/((s−2)²+9).

Do I need to memorise a transform table?

A short one: constants, tⁿ, e^(at), sin, cos, and the two shift theorems. Exams almost always supply the rest.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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