Questions: Solving Initial Value Problems with Laplace Transforms

Question 1

When solving y'' + 2y' + y = 0, y(0) = 3, y'(0) = −1 using Laplace transforms, at what point in the procedure do the initial conditions y(0) = 3 and y'(0) = −1 enter the calculation?

• A. After finding the general form of Y(s), as a separate substitution step
• B. At the inverse transform step, to select which particular solution to keep
• C. Automatically when the derivative rules L{y'} = sY − y(0) and L{y''} = s²Y − sy(0) − y'(0) are applied
• D. They are plugged in at the end to solve for arbitrary constants, as in the classical method

Answer: (C)

The structural advantage of the Laplace method is that initial conditions are incorporated automatically at the moment of transformation — they appear in the algebraic equation for Y(s) from the derivative rules. This is fundamentally different from the classical method, where you first find the general solution, then apply initial conditions as a separate step to determine constants. Options A, B, and D all describe classical-method thinking incorrectly applied to the Laplace approach.

Question 2

What algebraic operation in the s-domain corresponds to differentiation in the time domain under the Laplace transform?

• A. Division by s (differentiating becomes dividing)
• B. Multiplication by s (plus an initial condition term)
• C. Taking the derivative of Y(s) with respect to s
• D. Squaring Y(s) for second derivatives

Answer: (B)

L{y'(t)} = sY(s) − y(0): differentiation becomes multiplication by s (minus the initial condition). L{y''(t)} = s²Y(s) − sy(0) − y'(0): the second derivative becomes multiplication by s². This algebraic replacement of differentiation is precisely why the Laplace method converts an ODE into an algebraic equation. Option C (differentiating Y with respect to s) would correspond to multiplying y(t) by −t, not differentiating it.

Question 3

The Laplace transform method requires finding the homogeneous solution and a particular solution separately, then combining them and applying initial conditions.

• T. True
• F. False

Answer: False

This describes the classical undetermined-coefficients or variation-of-parameters approach. The Laplace method's key advantage is that it collapses all three stages — homogeneous solution, particular solution, and initial condition matching — into a single pass. You transform, solve algebraically for Y(s), and invert. The initial conditions are already embedded in Y(s) from the moment the derivative rules were applied.

Question 4

Partial fraction decomposition is the central algebraic step in the Laplace method because the functions in the inverse transform table — exponentials, sinusoids, polynomials — match the structural forms produced by partial fractions.

• T. True
• F. False

Answer: True

Once you have Y(s) as a ratio of polynomials, partial fractions decompose it into terms of the form A/(s−a), (As+B)/(s²+bs+c), and A/(s−a)^k. Each of these has a known inverse transform from the table. The match is exact by design: the Laplace method works because ODE solutions are built from exponentials, sinusoids, and polynomials — exactly the functions the table covers. Partial fractions is the bridge between solving for Y(s) and inverting it.

Question 5

Why is partial fraction decomposition the critical algebraic step in the Laplace transform method for solving IVPs, and what would happen if you couldn't decompose Y(s) into simpler terms?

Model answer: After solving for Y(s) algebraically, you typically have a ratio of polynomials that has no direct entry in the inverse transform table. Partial fraction decomposition breaks this into a sum of simple fractions — each of which matches a known inverse transform (exponential, sinusoidal, polynomial, or damped oscillation). Without this step, you cannot invert Y(s) back to y(t). If Y(s) couldn't be decomposed, you would be unable to find the time-domain solution using standard transform tables.

The Laplace method's power depends on the ability to invert Y(s). The inverse transform table only covers simple forms, not arbitrary rational functions. Partial fractions convert the algebraic solution into a sum of invertible pieces. This is why errors in partial fractions are the most common source of incorrect final answers — the transform and algebraic steps can be perfect, but a partial fraction mistake corrupts the inversion.