2 votes

Shows and ^(t^2) has no one transformed of Laplace

* Laplace's transformed * shows that e ^(t^2) has no one transformed of Laplace. Hint: Try to e ^(t^2-st) > et for t > s+1.

2voto

Anon User Points 0
We must start from the definition: L {f(t)} = ∫ (0 to ∞) e^(-st)\* f(t)\*dt; f(t) = e^(t^2); ∫ (0 to ∞) e^(-st)\* e^(t^2)\*dt; ∫ (0 to ∞) \[e^(t^2) / e^(st)\] \*dt ∫ (0 to ∞) e^(t^2 - st) \*dt; the diverger the function when t tends to ∞, has no resolution.

TheXpert.space

TheXpertSpace is an online community of people who give and receive advice.
You can check other people responses or create a new question if it's not already solved

Powered by: