Laplace Transforms and Applications

The Laplace transform is the board's favorite way to make a differential equation behave like ordinary algebra. It converts an unknown function of time, y(t), into a function of a new variable s, replaces derivatives with multiplication by s, and lets you solve a linear ODE by clearing fractions and reading an answer back out of a table. This lesson builds the transform from its integral definition, collects the standard pairs the MSTE paper leans on, then puts them to work inverting a rational transform and solving a second-order initial value problem all the way to a number.

The Laplace transform: definition

The Laplace transform of a function f(t) defined for t >= 0 is the integral

L{f(t)} = F(s) = integral from 0 to infinity of e^(-st) f(t) dt,

taken for values of s large enough that the integral converges. The transform trades the variable t for the variable s and, more importantly, trades calculus operations for algebraic ones. The whole strategy is a detour: instead of attacking a differential equation directly, you transform it into an algebraic equation in s, solve that by hand, and transform the answer back.

Two transforms fall straight out of the definition. For f(t) = 1, L{1} = integral from 0 to infinity of e^(-st) dt = [-(1/s) e^(-st)] from 0 to infinity = 0 - (-(1/s)) = 1/s, valid for s > 0. For f(t) = e^(at), L{e^(at)} = integral from 0 to infinity of e^(-st) e^(at) dt = integral of e^(-(s-a)t) dt = 1/(s - a), valid for s > a. You almost never re-derive these on the exam; you read them from a memorized table.

Transforms of the standard functions

The table below is the working core of the topic. Memorize the left half cold; the shifted forms in the last two rows follow from the first shifting theorem in the next section.

f(t)	L{f(t)} = F(s)
1	1/s
t^n (n a positive integer)	n! / s^(n+1)
e^(at)	1/(s - a)
sin(bt)	b / (s^2 + b^2)
cos(bt)	s / (s^2 + b^2)
e^(at) sin(bt)	b / ((s - a)^2 + b^2)
e^(at) cos(bt)	(s - a) / ((s - a)^2 + b^2)
u(t - a) (unit step)	e^(-as) / s

Worked example: Find L{t^3}. Using the power rule with n = 3, L{t^n} = n!/s^(n+1) gives 3!/s^(3+1) = 6/s^4. Contrast this with the common slip of forgetting the factorial, which would wrongly give 1/s^4.

Linearity and the first shifting theorem

The transform is linear: for constants a and b, L{a f(t) + b g(t)} = a F(s) + b G(s). That single fact lets you transform any sum term by term.

Worked example: Find L{4 e^(2t) - 5}. By linearity, L{4 e^(2t) - 5} = 4 L{e^(2t)} - 5 L{1} = 4/(s - 2) - 5/s. Nothing more is needed; each piece is a table entry scaled by its coefficient.

The first shifting theorem (the s-shift) states that multiplying f(t) by e^(at) shifts its transform by a:

L{e^(at) f(t)} = F(s - a),

where F(s) is the transform of f(t) alone. In words, wherever s appears in F(s), replace it by (s - a). This is exactly how the last two rows of the table are built. Starting from L{sin(bt)} = b/(s^2 + b^2) and multiplying by e^(at), every s becomes (s - a), giving L{e^(at) sin(bt)} = b/((s - a)^2 + b^2).

Worked example: Find L{e^(2t) t^2}. First transform the bare function, L{t^2} = 2!/s^3 = 2/s^3. Then shift s to (s - 2): L{e^(2t) t^2} = 2/(s - 2)^3.

Transform of a derivative

The property that makes the whole method work is the transform of a derivative. Integrating the definition by parts gives

L{f'(t)} = s F(s) - f(0),

and applying the rule twice gives the second derivative,

L{f''(t)} = s^2 F(s) - s f(0) - f'(0).

Read these carefully: differentiation in t becomes multiplication by s in the transform domain, and the initial conditions f(0) and f'(0) are folded in automatically. That is why you never need a separate step to fit the constants of integration; the initial values enter the algebra from the start.

Inverse transforms by partial fractions

Solving the algebra in s hands you a rational function of s, and the inverse transform L^(-1) turns it back into a function of t. The one skill that unlocks almost every inverse on the paper is partial fractions: factor the denominator, split the fraction into pieces that each match a table entry, then invert term by term. When the denominator is an irreducible quadratic, complete the square and match the shifted sine and cosine forms.

Worked example (inverting a rational transform): An RLC circuit produces the voltage transform V(s) = (s + 3) / (s^2 + 2s + 5) volts-seconds; find v(t) and its value at t = 1 s. Complete the square in the denominator: s^2 + 2s + 5 = (s + 1)^2 + 4, so the shift is a = -1 and b = 2. Split the numerator to match the shifted cosine and sine forms by writing s + 3 = (s + 1) + 2:

V(s) = (s + 1) / ((s + 1)^2 + 2^2) + 2 / ((s + 1)^2 + 2^2).

The first piece is the transform of e^(-t) cos(2t). The second piece matches b/((s - a)^2 + b^2) with b = 2, which is the transform of e^(-t) sin(2t). Therefore v(t) = e^(-t) cos(2t) + e^(-t) sin(2t) = e^(-t)(cos 2t + sin 2t) volts. At t = 1 s, e^(-1) = 0.3679, cos(2 rad) = -0.4161, and sin(2 rad) = 0.9093, so v(1) = 0.3679 (-0.4161 + 0.9093) = 0.3679(0.4932) = 0.181 V, about 0.18 V.

Solving a linear ODE with initial conditions

Now the pieces combine. To solve a linear constant-coefficient ODE, transform every term, substitute the initial conditions through the derivative rules, solve the resulting algebraic equation for Y(s), and invert.

Worked example (a second-order IVP): A damped spring-mass system obeys x'' + 3x' + 2x = 0 with x(0) = 1 m and x'(0) = 0 m/s; find the displacement x(t) and its value at t = 1 s. Transform each term, writing X(s) for L{x(t)}:

L{x''} = s^2 X - s x(0) - x'(0) = s^2 X - s, and L{x'} = s X - x(0) = s X - 1.

Substituting into the ODE, (s^2 X - s) + 3(s X - 1) + 2X = 0, which collects to (s^2 + 3s + 2) X = s + 3. Solve for X and factor the denominator:

X(s) = (s + 3) / (s^2 + 3s + 2) = (s + 3) / ((s + 1)(s + 2)).

Decompose by partial fractions, (s + 3)/((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2), so s + 3 = A(s + 2) + B(s + 1). Setting s = -1 gives 2 = A, and setting s = -2 gives 1 = -B, so B = -1. Then X(s) = 2/(s + 1) - 1/(s + 2), which inverts term by term to

x(t) = 2 e^(-t) - e^(-2t) meters.

Check the initial conditions: x(0) = 2 - 1 = 1 m, and x'(t) = -2 e^(-t) + 2 e^(-2t) gives x'(0) = -2 + 2 = 0 m/s, both correct. At t = 1 s, x(1) = 2(0.3679) - 0.1353 = 0.7358 - 0.1353 = 0.600 m, about 0.60 m.

The unit step function and its transform

Board problems that switch a load or voltage on at a set time use the unit step (Heaviside) function u(t - a), which is 0 for t < a and 1 for t >= a. It turns a source on at t = a.

Its transform is L{u(t - a)} = e^(-as)/s. More generally, the second shifting theorem (the t-shift) handles a whole function delayed to start at t = a: L{f(t - a) u(t - a)} = e^(-as) F(s). The exponential factor e^(-as) is the transform-domain fingerprint of a time delay, so an e^(-as) in a transform always signals a step or a delayed source.

Worked example: Find L{u(t - 2)}. Reading L{u(t - a)} = e^(-as)/s with a = 2 gives L{u(t - 2)} = e^(-2s)/s directly.

Exam-day strategy

• Work from a memorized transform table; the pairs for 1, t^n, e^(at), sin(bt), and cos(bt) cover most items, and the two e^(at) rows are just the sine and cosine rows with s shifted to (s - a).
• Remember the factorial in L{t^n} = n!/s^(n+1); a missing factorial is the single most common table error.
• For an inverse whose denominator is an irreducible quadratic, complete the square first, then split the numerator into an (s - a) part and a constant part to match the shifted cosine and sine forms.
• When transforming a derivative, write L{f''} = s^2 F(s) - s f(0) - f'(0) with the initial conditions already in place; there is no separate step to solve for a constant afterward.
• Factor the denominator and use partial fractions before inverting; the cover-up shortcut (set s to each root) gives the constants in one line.
• Treat any e^(-as) factor as a flag for the unit step or a delayed source, and read the delay time a straight out of the exponent.
• Always verify the solution against both initial conditions; substituting t = 0 into y(t) and y'(t) is a fast check that catches a dropped sign in the partial fractions.