§ 01The Big Idea — Turning Calculus into Algebra
Differentiation is hard. Algebra is easy. The Laplace transform is a machine that converts the hard problem into the easy one: then converts the answer back.
Every technique covered so far, separation of variables, integrating factors, the characteristic equation, undetermined coefficients: requires you to guess a solution form, differentiate, substitute, and match coefficients. The Laplace transform sidesteps all of this. Instead of working in the time domain where the unknown is a function y(t), you work in the s-domain where the unknown is an algebraic expression Y(s). Differentiation becomes multiplication by s. The ODE becomes a polynomial equation in s. Solve it. Translate back.
hard — involves derivatives
ℒ{ }
easy — no derivatives
for Y(s)
algebraic expression
ℒ−1{ }
the answer you wanted
This approach has two killer advantages over other methods. First, initial conditions are built into the transform automatically: you do not solve the general solution and then apply conditions afterwards. Second, the Laplace method handles discontinuous or impulsive forcing functions, like a sudden voltage spike or a bang on a mechanical system, that make undetermined coefficients completely helpless. These are modelled by the Heaviside step function and the Dirac delta, which we cover below.
§ 02The Definition
The Laplace transform is an improper integral that converts a function of t into a function of a complex parameter s.
A few immediate observations:
The variable s. The parameter s must be large enough to make the integral converge. For most functions in this course, s just needs to be a positive real number sufficiently large. In advanced work s is complex, but we will not need that here.
The factor e−st. This exponential damping factor is what makes the integral converge. For a function that grows no faster than eat, the integral converges for all s > a. The smallest such value is called the abscissa of convergence.
Notation. We write ℒ{f(t)} = F(s). Capital letters denote the transform. Lowercase letters denote time-domain functions. This convention is universal and you should use it consistently.
Computing ℒ{1} from the Definition
Let f(t) = 1 for all t ≥ 0. Then:
= limb→∞ [−(1/s)e−st]0b
= limb→∞ [−(1/s)e−sb + (1/s)e0]
= 0 + 1/s (for s > 0)
ℒ{1} = 1/s
Computing ℒ{eat} from the Definition
= [−1/(s−a) · e−(s−a)t]0∞ (for s > a)
= 0 + 1/(s−a)
ℒ{eat} = 1/(s−a) for s > a
This single result is responsible for roughly half of all the Laplace transforms you will ever compute. Every entry in the transform table that involves eat is a variant of this formula.
Computing ℒ{sin(ωt)} and ℒ{cos(ωt)}
These are found via integration by parts twice (or more elegantly via Euler's formula eiωt = cos ωt + i sin ωt and then taking real and imaginary parts). The results are:
ℒ{cos(ωt)} = s/(s2 + ω2)
Notice the symmetry: the transform of sin has ω on top; the transform of cos has s on top. If you ever mix them up, remember that cos(0) = 1 and sin(0) = 0, and check that the inverse at t = 0 gives the right value.
Computing ℒ{tn}
Repeated integration by parts (or an induction argument) gives:
For n = 0 this recovers ℒ{1} = 1/s. For n = 1: ℒ{t} = 1/s2. For n = 2: ℒ{t2} = 2/s3. Each additional power of t adds one to the power of s in the denominator and one factorial factor on top.
§ 03Linearity and the Derivative Theorem
Two properties make Laplace transforms practical: linearity (which handles sums) and the derivative theorem (which is why the method converts ODEs into algebra).
Linearity
The Laplace transform is a linear operator:
= αF(s) + βG(s)
This follows directly from the linearity of integration. It means you can transform sums term by term and pull constants outside the transform: just like integration itself.
The First Derivative Theorem — The Heart of the Method
This is the most important result on the whole page. Integrate by parts with u = e−st and dv = f′(t)dt:
= [e−stf(t)]0∞ − ∫0∞ (−s)e−st f(t) dt
= 0 − f(0) + s∫0∞ e−stf(t) dt
ℒ{f′(t)} = sF(s) − f(0)
Read this carefully: the transform of the derivative f′(t) equals s times the transform F(s) minus the initial value f(0). Differentiation in the t-domain becomes multiplication by s in the s-domain. That is why ODEs become algebraic equations.
Applying the result a second time (replace f by f′ in the formula above):
ℒ{y″} = s2Y(s) − s⋅y(0) − y′(0)
These two formulas, which you must memorise, are all you need to convert any constant-coefficient second-order IVP into an algebraic equation. Notice how the initial conditions y(0) and y′(0) appear automatically, embedded inside the s-domain equation.
The First Shift Theorem (s-Shifting)
If ℒ{f(t)} = F(s), then:
Proof: substitute s − a for s in the definition integral. The eat factor in the integrand combines with e−st to give e−(s−a)t, which is exactly the integral for F(s−a). This theorem lets you handle transforms like ℒ{e2tsin(3t)} without computing any new integral from scratch: it equals the sin transform with s replaced by s−2, giving 3/((s−2)2+9).
The Second Shift Theorem (t-Shifting) and the Heaviside Function
The Heaviside step function uc(t) is defined as 0 for t < c and 1 for t ≥ c. It models a switch that turns on at time c. Its transform is:
ℒ{uc(t)⋅f(t−c)} = e−csF(s)
In words: a time-delay of c in the t-domain corresponds to multiplication by e−cs in the s-domain. This is what makes Laplace transforms invaluable for piecewise and switched forcing functions in engineering.
§ 04Complete Transform Table
These are the entries you will need for the vast majority of undergraduate ODE problems. The first column is the time-domain function f(t); the second is its Laplace transform F(s) = ℒ{f(t)}.
| f(t) (t ≥ 0) | F(s) = ℒ{f(t)} | Conditions / notes |
|---|---|---|
| ▶ Constants and Powers | ||
| 1 | 1/s | s > 0 |
| t | 1/s2 | s > 0 |
| t2 | 2/s3 | s > 0 |
| tn | n!/sn+1 | s > 0, n = 0,1,2,… |
| t1/2 | π1/2/(2s3/2) | Uses Γ(3/2) = π1/2/2 |
| ▶ Exponentials | ||
| eat | 1/(s−a) | s > a |
| t⋅eat | 1/(s−a)2 | s > a |
| tneat | n!/(s−a)n+1 | s > a, n = 0,1,2,… |
| ▶ Trigonometric | ||
| sin(ωt) | ω/(s2+ω2) | s > 0 |
| cos(ωt) | s/(s2+ω2) | s > 0 |
| t⋅sin(ωt) | 2ωs/(s2+ω2)2 | s > 0 |
| t⋅cos(ωt) | (s2−ω2)/(s2+ω2)2 | s > 0 |
| ▶ Damped Oscillations (First Shift) | ||
| eatsin(ωt) | ω/((s−a)2+ω2) | s > a |
| eatcos(ωt) | (s−a)/((s−a)2+ω2) | s > a |
| ▶ Hyperbolic | ||
| sinh(at) | a/(s2−a2) | s > |a| |
| cosh(at) | s/(s2−a2) | s > |a| |
| ▶ Special Functions | ||
| uc(t) (Heaviside) | e−cs/s | c ≥ 0 |
| δ(t) (Dirac delta) | 1 | unit impulse at t = 0 |
| δ(t−c) | e−cs | impulse at t = c |
| ▶ Derivative Rules | ||
| f′(t) | sF(s) − f(0) | |
| f″(t) | s2F(s) − sf(0) − f′(0) | |
| f(n)(t) | snF(s) − sn−1f(0) − ⋯ − f(n−1)(0) | |
| ▶ Other Operational Rules | ||
| t⋅f(t) | −F′(s) | differentiate the transform |
| f(t)/t | ∫s∞ F(u) du | integrate the transform |
| (f*g)(t) (convolution) | F(s)⋅G(s) | see §06 |
§ 05Solving IVPs — The Four-Step Method
Every Laplace transform solution follows the same four steps. Master the pattern and you can solve any constant-coefficient linear IVP.
2. Use derivative rules to replace y′, y″ with Y(s) — initial conditions appear automatically
3. Solve the resulting algebra for Y(s)
4. Find y(t) = ℒ−1{Y(s)} using the table (and partial fractions if needed)
Partial Fractions in the s-Domain
After Step 3, Y(s) is almost always a rational function: a ratio of polynomials in s. Step 4 requires writing this as a sum of simpler fractions that match table entries. The rules are exactly those from the partial fractions section of integration, applied to the variable s instead of x.
The three cases that arise most often:
Repeated linear: A/(s−a) + B/(s−a)2 ⇋ Aeat + Bteat
Irreducible quad.: (As+B)/(s2+ω2) ⇋ A⋅cos(ωt) + (B/ω)⋅sin(ωt)
When the quadratic is shifted (of the form (s−a)2 + ω2) complete the square first, then use the damped oscillation entries in the table.
§ 06Worked Examples
Twelve fully solved problems: forward transforms, inverse transforms, first-order IVPs, second-order IVPs, repeated roots, complex roots, and a Heaviside forcing function.
Find ℒ{3 − 2t + 4e5t − 7sin(2t)}.
Find ℒ{e2tcos(3t)}.
Find ℒ−1{(2s+8)/((s+1)(s+3))}.
Find ℒ−1{(s+3)/(s2+6s+13)}.
Solve: y′ + 3y = 6et, y(0) = 1
Solve: y″ − 5y′ + 6y = 0, y(0) = 1, y′(0) = 2
Solve: y″ + 4y = 0, y(0) = 3, y′(0) = −2
Solve: y″ + 2y′ + y = e−t, y(0) = 0, y′(0) = 0
Solve: y″ + 9y = 18cos(3t), y(0) = 0, y′(0) = 0
Solve: y′ + 2y = uπ(t), y(0) = 0 where uπ is the Heaviside function that switches on at t = π.
Solve: y″ + 3y′ + 2y = δ(t−1), y(0) = 0, y′(0) = 0
Use the convolution theorem to find ℒ−1{1/[s(s2+1)]}.
Cross-check: partial fractions gives 1/s − s/(s²+1), so ℒ−1 = 1 − cos(t). ✓
§ 07The Convolution Theorem
Convolution lets you invert a product of transforms without using partial fractions, and it has deep physical meaning as a weighted running average.
The convolution of two functions f and g is defined as:
Note that f * g = g * f (convolution is commutative), so you can choose whichever ordering makes the integral easier to compute.
equivalently: ℒ−1{F(s)⋅G(s)} = (f*g)(t)
The theorem tells you: multiplication in the s-domain corresponds to convolution in the t-domain. This is analogous to how the Fourier transform converts convolution into pointwise multiplication in the frequency domain, in fact it is the same fundamental idea.
When to use convolution. If partial fractions would be messy but you can recognise F(s) and G(s) as individual table entries, the convolution integral is often cleaner. It also appears naturally in the analysis of linear systems: if H(s) is the transfer function of a system and X(s) is the input, then Y(s) = H(s)X(s), so the output y(t) is the convolution of the impulse response h(t) with the input x(t).
§ 08Common Mistakes
§ 09Practice Quiz — 10 Questions
Test your mastery of forward transforms, inverse transforms, derivative rules, and solving IVPs.
The Laplace transform is defined as ℒ{f(t)} = ∫0∞ f(t) ⋅ _____ dt. What fills the blank?
ℒ{e7t} = 1/(s − k). What is k?
If Y(s) = ℒ{y(t)}, which expression equals ℒ{y″(t)}?
ℒ{t3} = k/s4. What is k?
- ℒ{t3} = 3!/s4 = 6/s4
What is ℒ−1{4/(s2+16)}?
- ℒ{sin(ωt)} = ω/(s²+ω²). Here ω = 4.
- 4/(s²+16) = ℒ{sin(4t)}, so ℒ−1{4/(s²+16)} = sin(4t).
ℒ{e−2tsin(5t)} = 5/((s+k)2+25). What is k?
What is ℒ{5δ(t) + 3et}?
- ℒ{δ(t)} = 1, so ℒ{5δ(t)} = 5
- ℒ{et} = 1/(s−1), so ℒ{3et} = 3/(s−1)
- Answer: 5 + 3/(s−1)
Apply Laplace to y′ − 4y = 0, y(0) = 3. The solution is y(t) = Ce4t. What is C?
- Transform: sY − 3 − 4Y = 0
- Y(s−4) = 3 → Y = 3/(s−4)
- Invert: y(t) = 3e4t, so C = 3
The convolution theorem states that ℒ{(f*g)(t)} equals:
Solve y″ + y = 0, y(0) = 0, y′(0) = 4. The solution is y(t) = A⋅sin(t). What is A?
- ℒ{y″} + ℒ{y} = 0: (s²Y − 0⋅s − 4) + Y = 0
- Y(s²+1) = 4 → Y = 4/(s²+1)
- ℒ{sin(t)} = 1/(s²+1), so Y = 4⋅ℒ{sin(t)}
- y(t) = 4sin(t), so A = 4
§ 10What to Study Next
Laplace transforms round off the core first and second-order ODE toolkit. From here, two natural directions open up:
- Variation of Parameters: a completely general method for non-homogeneous second-order ODEs when undetermined coefficients fails.
- Power Series, series solutions to ODEs with variable coefficients (Frobenius method), the tool for equations like Bessel's and Legendre's.
- Partial Fractions: sharpen this technique as it is the bottleneck in almost every Laplace inversion.
- Partial Derivatives, prerequisite for partial differential equations (PDEs), where Laplace transforms in time are used alongside Fourier series in space.