1
State the transform and compute the first entries of the table
Write the defining integral, compute ℒ[1] and ℒ[e^(at)] from truncated integrals, and say for which s each result holds.
Course · Differential Equations
Laplace Transforms
Every differential equation so far has been solved in the time domain: guess a form, match coefficients, then fit the constants to the starting values. The Laplace transform moves the whole problem to a different variable, s, where y′ becomes s·Y − y(0) and y″ becomes s²·Y − s·y(0) − y′(0). The equation turns into (a·s² + b·s + c)·Y = something, a line of algebra, and the starting values are already inside it. What is left is a rational function to take apart into the pieces of a short table. This course builds the table, proves the derivative rule from the product rule, solves initial value problems end to end, and then reads what the poles of the answer say about the long run. Every transform and every coefficient in the labs is an exact fraction, and every answer is checked by substituting it back.
ℒ[f](s) = ∫₀^∞ e^(−st)·f(t) dt
ℒ[e^(at)] = 1/(s − a)
ℒ[y′] = s·Y − y(0)
(a·s² + b·s + c)·Y = initial terms + ℒ[q]
partial fractions, then the table backwards
poles = characteristic roots
ℒ[u(t − c)·f(t − c)] = e^(−cs)·F(s)
What this course is about
- Subject
- Differential Equations
- Lessons
- 8
- Format
- Interactive
Overview
What you will be able to do
By the end you can transform a signal built from powers, exponentials and sinusoids, solve a constant-coefficient initial value problem by algebra, invert the answer by partial fractions, and check it by substitution.
2
Transform a signal with linearity and the table
Split a signal into table entries, scale and add their transforms, and combine the result into one fraction in lowest terms.
3
Apply the derivative rule
Write ℒ[y′] and ℒ[y″] with their initial values, and verify the rule on a signal by transforming its derivative directly.
4
Solve an initial value problem by algebra
Transform both sides, solve for Y(s), invert through partial fractions and verify the result by substitution.
5
Decompose a rational function exactly
Handle linear, repeated and irreducible quadratic factors and invert each piece.
6
Read stability and switched forcing from the transform
Identify poles with characteristic roots, state the stability criterion, and solve an equation whose forcing switches on at a stated time.
Lessons
8 lessons
The transform first, because everything else is a use of it: its definition, its table and its rule for derivatives. Solving comes next, as one procedure and then as the two pieces of it that need care, the partial fractions and the poles. The course ends with forcing that switches on at a time, and with one equation solved by two methods that have to agree.
The Laplace TransformWeight a signal by e^(−st), add up everything from t = 0 onward, and the result is a function of s that, for these signals, is a fraction.Not marked
Linearity and the TableThe transform of a sum is the sum of the transforms, constants come straight out, and five table entries then cover every signal this course uses.Not marked
The Transform of a Derivativeℒ[y′] is s times Y less the starting value, and that starting value is what puts the initial conditions into the equation.Not marked
Solving an Initial Value ProblemTransform both sides, solve for Y by algebra, invert with partial fractions, and the initial values are already in the answer.Not marked
Partial Fractions, ExactlyA fraction in s is taken apart into one term for every power of every factor, and each term is a table entry read backwards.Not marked
The Transfer Function and Its PolesFrom rest, the response is a fixed fraction times the transform of the forcing, and the long run is read from where that fraction's denominator vanishes.Not marked
The Unit Step and Switched ForcingA forcing that switches on at t = c transforms to e^(−cs) times an ordinary fraction, and the solution is that fraction's inverse, shifted to start at c.Not marked
Three Methods, One EquationUndetermined coefficients, the Laplace transform and, for a first-order equation, the integrating factor all return the same function, and substitution is the referee.Not marked
Recommended background
The previous course, Systems and the Phase Plane, for the habit of reading eigenvalues and roots as the numbers that decide long-run behaviour, together with Second-Order Linear Equations for the characteristic equation and the fitting of constants, Accumulation and the Integral for the integral as accumulated change, and Algebra's Rational and Radical Expressions for adding and reducing fractions; partial fractions are not assumed, since that course leaves them out, and are built here from scratch.
Practice suggestions
Keep the two variables apart. A signal lives in t, its transform lives in s, and a figure belongs to one of them. The labs print the transform as a fraction in s and the solution as a sum of exponentials in t, and the status line says which is which.
Check every answer by substitution. The lab reports the residual of its solution in the original equation, and residual 0 is the claim that the algebra and the inversion were both right. Do the same check by hand on one preset in each lesson.
Do the algebra before reading the lab. The transformed equation, Y(s) and the partial fractions are all short enough to write out, and a lab that agrees with your line is worth more than one you copied from.
Say which figures are exact. The transforms, the coefficients and the poles are exact fractions or exact surd-free roots. The truncated integrals that the first lesson shows are rounded and printed with ≈, because they are evidence for a claim and not the claim.
What it does not cover
The Dirac delta, which needs distributions, and the convolution theorem. Both are standard in a full treatment of the transform; here the forcing is a polynomial, an exponential, a sinusoid or a switch, and the response to an impulse is not needed.
Transforms of periodic functions, the inversion integral and the region of convergence as a theory. The lessons state the half-line of s where each integral converges and use it; they do not develop the complex analysis that explains it.
Irrational frequencies. A denominator such as s² + 2 has the pole pair ±i√2, and its inverse has cos(√2·t) in it. The lab refuses that case and names the factor rather than print a surd inside a cosine.
Systems of equations by transform. The previous course solves them with eigenvalues and eigenvectors, and the transform here is applied to a single equation of first or second order.
Equations with variable coefficients, where the transform gives a differential equation in s and not an algebraic one. The method works because the coefficients are constant.