Questions: Laplace Transform: Definition and Properties

Question 1

What does the Laplace transform derivative rule L{f'(t)} = sF(s) - f(0) accomplish that makes it powerful for solving initial value problems?

• A. It eliminates the need to find F(s) by directly computing the answer in the time domain
• B. It converts differentiation in t into multiplication by s, incorporating the initial condition algebraically so the ODE becomes an algebraic equation in s
• C. It shows that all derivatives have the same Laplace transform, simplifying the computation
• D. It converts an algebraic equation back into a differential equation so it can be solved by standard methods

Answer: (B)

The derivative rule's entire power is in this conversion: instead of solving a differential equation, you solve an algebraic equation in F(s). The initial condition f(0) appears as part of the algebra — you don't need to impose it as a separate step after solving. This is why Laplace transforms are especially valuable for IVPs: initial conditions are baked in automatically.

Question 2

A student has the ODE f'' + 4f' + 3f = e^{-2t} with f(0) = 1, f'(0) = 0. After applying the Laplace transform, what kind of equation does she need to solve?

• A. A new second-order ODE in F(s)
• B. An algebraic equation in F(s) where the initial conditions already appear as constants
• C. An integral equation that requires numerical methods
• D. The same ODE, now written in the s-variable instead of t

Answer: (B)

Applying the transform converts every derivative into multiplication by s (with initial conditions as constants) and every function into its transform. The result is a purely algebraic equation in F(s) — no differential equation remains, just algebra. The initial conditions f(0)=1 and f'(0)=0 appear in the algebra automatically, without any separate step to impose them.

Question 3

The Laplace transform L{e^{at}f(t)} = F(s - a) means that multiplying a function by an exponential in t shifts the argument of its transform in s.

• T. True
• F. False

Answer: True

This is the s-shifting (first shifting) theorem. Multiplying f(t) by e^{at} in the time domain shifts the argument of F(s) from s to s - a in the frequency domain. This is useful for handling forcing functions with exponential growth or decay — you can use a known transform F(s) and simply replace s with s - a. The reverse direction (t-shifting) handles functions that switch on at some time c via the unit step function.

Question 4

The Laplace transform method is most advantageous over direct methods when the forcing function in an ODE is a smooth, continuous function like a polynomial.

• T. True
• F. False

Answer: False

The Laplace transform's greatest advantage over direct methods (variation of parameters, undetermined coefficients) appears with discontinuous forcing functions like step functions and impulses. For smooth forcing functions, direct methods often work just as easily. The t-shifting theorem and the Heaviside step function make the Laplace transform the natural tool when a forcing function 'switches on' at some time c — a case that is awkward and messy to handle any other way.

Question 5

Explain why the Laplace transform converts a differential equation with initial conditions into an algebraic equation, and why this is more than just a computational shortcut.

Model answer: The derivative rule L{f'(t)} = sF(s) - f(0) replaces differentiation with multiplication by s (plus an algebraic term from the initial condition). A second derivative L{f''(t)} = s²F(s) - sf(0) - f'(0) reduces the order again. Applying the transform to a whole ODE converts every term into an algebraic expression in F(s), including the initial conditions, which appear as constants rather than side conditions. The result is an algebraic equation solvable for F(s), after which inversion recovers f(t).

This is conceptually important because it changes the mathematical domain of the problem. Rather than searching for a function f(t) that satisfies a differential equation, you search for an algebraic function F(s), then recover f(t) via inverse transform. Initial conditions are absorbed into the algebra — they don't need to be imposed after finding a general solution, which eliminates an entire step and handles discontinuous forcing functions cleanly.