Find the Laplace transform of f(t) = 5.
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Answer:
Use the table entry for a constant.
In this topic you solve differential equations with algebra. You transform the unknown y(t) into a function Y(s) using a table. The equation becomes an algebra problem. You solve for Y(s), split it into simple pieces, and read y(t) back from the table. Recognition cue: the problem gives starting values like y(0) and asks you to use the Laplace transform, or a force switches on at a set time. You will practice: Transform table and linearity, First shifting theorem, Transforms of derivatives, Partial fraction inversion.
Find the Laplace transform of f(t) = 5.
Answer:
Use the table entry for a constant.
Find the Laplace transform of f(t) = e^(3t).
Answer:
Use the exponential entry with a = 3.
Find the Laplace transform of f(t) = t².
Answer:
Use the power entry with n = 2.
Find the Laplace transform of f(t) = sin(4t).
Answer:
Use the sine entry with b = 4.
A sensor outputs the voltage v(t) = 2t + 7 volts. Find its Laplace transform V(s).
Answer:
Transform each term and keep the constants in front.
Find the Laplace transform of f(t) = e^(-t) cos(2t).
Answer:
Start from the cosine transform. The factor e^(-t) replaces s by s + 1.
Find the inverse Laplace transform of F(s) = 3/(s - 5).
Answer:
Match the exponential entry with a = 5.
Find the inverse Laplace transform of F(s) = 8/(s² + 16).
Answer:
The denominator says b = 4, so the top must be 4. Factor out 2.
Let Y(s) be the Laplace transform of y(t), and suppose y(0) = 2. Write the Laplace transform of y' in terms of Y(s) and s.
Answer:
Apply the derivative rule directly.
A bacteria population grows by P' = 3P, with P(0) = 4 million cells. Use the Laplace transform to find P(t) in millions.
Answer:
Write Y(s) for the transform of P. Transform, solve for Y, invert.
Check: P' = 12e^(3t) = 3P and P(0) = 4.
Compute the Laplace transform of f(t) = e^{2t} directly from the definition L{f}(s) = ∫₀^∞ e^{−st} f(t) dt. State the values of s for which the integral converges.
Answer:
Combine the exponentials and integrate on [0, T].
For s > 2 the exponential term vanishes as T grows, leaving 1/(s − 2). For s ≤ 2 the integral diverges.
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