Laplace Transforms
Differential Equations · Section 6.1 · generated practice set
Working the transform table instead of the definition, plus the first shifting theorem.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Using the Laplace transform table, find $\mathcal{L}\left\{e^{2t}\sin(2t)\right\}$.
Solution
- Match the function to a table entry rather than integrating from the definition.
- Key entries: $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$, $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$, $\mathcal{L}\{\sin bt\} = \frac{b}{s^2+b^2}$, $\mathcal{L}\{\cos bt\} = \frac{s}{s^2+b^2}$.
- Multiplying by $e^{at}$ in $t$ shifts $s \to s - a$ in the transform (first shifting theorem).
- Result: $\mathcal{L}\left\{e^{2t}\sin(2t)\right\} = \dfrac{2}{(s - 2)^2 + 4}$.
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Method
Compute from the definition $\mathcal{L}\{f\} = \int_0^\infty e^{-st}f(t)\,dt$ only when asked; otherwise use the table. The transform is linear, and multiplying by $e^{at}$ in $t$ shifts $s \to s - a$.
Definitions and theorems
Worked example
Find $\mathcal{L}\{t^2e^{-3t}\}$.
- Start from $\mathcal{L}\{t^2\} = \frac{2}{s^3}$.
- Apply the shift $s \to s+3$.
- $\mathcal{L}\{t^2e^{-3t}\} = \frac{2}{(s+3)^3}$.
Common mistakes
- Shift direction. $e^{at}$ sends $s \to s-a$, so $e^{-3t}$ gives $(s+3)$.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.