Questions: Laplace Transform of Derivatives and Integrals

Question 1

What does the formula L{f'(t)} = sF(s) − f(0) accomplish when applied to an ODE?

• A. It replaces differentiation with division by s, simplifying integration
• B. It converts differentiation in the time domain into multiplication by s in the s-domain, with initial conditions encoded algebraically
• C. It eliminates the need for initial conditions by shifting them to a boundary term
• D. It converts the ODE into a partial differential equation in s and t

Answer: (B)

The key transformation is that differentiation (an operation requiring calculus) becomes multiplication by s (a purely algebraic operation) in the s-domain. The initial condition f(0) is subtracted as a constant term. This means a differential equation — which requires solving for a function — becomes a polynomial equation in Y(s) that can be solved by algebra alone, then inverted.

Question 2

For the ODE y'' + 3y' + 2y = eᵗ with y(0) = 0 and y'(0) = 1, what does the left side become after taking the Laplace transform?

• A. (s² + 3s + 2)Y(s)
• B. (s² + 3s + 2)Y(s) − 1
• C. (s² + 3s + 2)Y(s) − s
• D. s²Y(s) + 3sY(s) + 2Y(s) − s − 3

Answer: (B)

Applying the formulas: L{y''} = s²Y − sy(0) − y'(0) = s²Y − 0 − 1 = s²Y − 1. L{y'} = sY − y(0) = sY − 0 = sY. L{2y} = 2Y. Summing: s²Y − 1 + 3sY + 2Y = (s² + 3s + 2)Y − 1. The −1 comes entirely from the initial condition y'(0) = 1. Option A forgets this term; options C and D incorrectly apply the initial conditions.

Question 3

The Laplace transform of f'(t) is derived using integration by parts.

• T. True
• F. False

Answer: True

L{f'(t)} = ∫₀^∞ f'(t)e^{−st} dt is evaluated by integration by parts with u = e^{−st} and dv = f'(t)dt. This yields [f(t)e^{−st}]₀^∞ + s∫₀^∞ f(t)e^{−st} dt = −f(0) + sF(s). The integration by parts is the entire derivation — the formula is not a definition but a theorem proved this way.

Question 4

Differentiating twice gives L{f''(t)} = s²F(s) − sf(0) − f'(0), which incorporates both initial conditions y(0) and y'(0).

• T. True
• F. False

Answer: True

Applying the first-derivative formula twice: L{f''} = L{(f')'} = sL{f'} − f'(0) = s(sF(s) − f(0)) − f'(0) = s²F(s) − sf(0) − f'(0). Each differentiation 'peels off' one initial condition. This is precisely why Laplace transforms are the natural tool for initial value problems: the initial conditions are automatically incorporated into the transformed equation.

Question 5

Why do initial conditions appear in the Laplace transform of derivatives, and why is this advantageous for solving initial value problems?

Model answer: Initial conditions appear because of the boundary term that arises from integration by parts: [f(t)e^{−st}]₀^∞ evaluates to −f(0) at the lower limit. Higher derivatives peel off successive initial conditions (f(0), f'(0), etc.) as additional terms. This is advantageous because, instead of first finding the general solution and then fitting constants to initial conditions in a separate step, the initial conditions are baked directly into the transformed algebraic equation — the solution automatically satisfies them.

This is the core payoff of the method. In standard ODE techniques (undetermined coefficients, variation of parameters), you find a general solution C₁y₁ + C₂y₂ + yₚ and then solve a system of equations to find C₁ and C₂ from initial conditions. The Laplace approach collapses these two steps: the transformed equation already incorporates the initial conditions as constants, so solving for Y(s) directly gives the specific (not general) solution.