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The Laplace transform of $f(t)=e^{2t}\sin(5t)u(t)$ is

1. $\frac{5}{s^2-4s+29}$
2. $\frac{5}{s^2+5}$
3. $\frac{s-2}{s^2-4s+29}$
4. $\frac{5}{s+5}$

Correct answer: $\frac{5}{s^2-4s+29}$

Solution

For $\sin(5t)$, the Laplace transform is $\frac{5}{s^2+25}$. Using the exponential shift property, $\mathcal{L}\{e^{2t}\sin(5t)\}=\frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}$.

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