Question

Find the Laplace transform for
f(t) = d/dt.(e⁻ᵃᵗ sin ωt)

Answer 1

The Laplace transform of f(t) = d/dt. (e^-at sin(ωt)) is -a/(s + a) * ω / (s^2 + ω^2) + ω^2 / ( (s + a)^2 + ω^2 ).

Step-by-step explanation:

The Laplace transform for f(t) = d/dt. (e^-at sin(ωt)) can be found using the property that the Laplace transform of the derivative of a function is sF(s) - f(0), where F(s) is the Laplace transform of the function. Let's find the Laplace transform step by step:

1. Find the Laplace transform of e^-at. The Laplace transform of e^-at is 1/(s+a).
2. Find the Laplace transform of sin(ωt). The Laplace transform of sin(ωt) is ω / (s^2 + ω^2).
3. Take the derivative of e^-at sin(ωt). The derivative of e^-at sin(ωt) is -ae^-at sin(ωt) + ωe^-at cos(ωt).
4. Apply the Laplace transform property for the derivative. The Laplace transform of -ae^-at sin(ωt) is -a/(s + a) * ω / (s^2 + ω^2).
5. The Laplace transform of ωe^-at cos(ωt) is ω^2 / ( (s + a)^2 + ω^2 ).
6. Add the results from steps 4 and 5 to get the Laplace transform for f(t) = d/dt. (e^-at sin(ωt)).

Therefore, the Laplace transform for f(t) = d/dt. (e^-at sin(ωt)) is -a/(s + a) * ω / (s^2 + ω^2) + ω^2 / ( (s + a)^2 + ω^2 ).