Basic Transforms
This function has a Laplace transform 1/s ?
What is 1?
L−1{1/s}
1
This term appears in L{y′}
What is −y(0)?
1/(s + 2)
Already in its simplest form
L{u(t−3)}
(e^−3s)/s
L{t} ?
1/(s^2)
L−1{1/s^2}
t
y′=y,y(0)=1
e^t
3/{(s + 1)(s + 2)}
3/(s-1) − 3/ (s + 2)
What does u(t−2) OR (t−2)u(t-2) represent?
Function turns on at t = 2
L{e^3t}
1/ (s-3)
L−1{1/ (s−4)}
e^4t
y′+y=0, y(0)=2
2e^−t
4/ {(s+2)(s+3)}
4/(s+2) − 4/(s+3)
L{u(t−1)(t−1)}
(e^−s) / s^2
L{cos(2t)}
s/(s^2 + 4)
L−1{3/(s^2 + 9}
sin(3t)
y′′+y=0, y(0)=0, y′(0)=1
sin(t)
s/(s^2 + 4)
Already standard (cos form)
L{u(t−2)e^(3(t−2))}
(e^−2s) / (s−3)
L{2t^2 + 3e^t − sin(2t)}
4/s^3 + 3/(s−1) − 2/(s^2 + 4)
L−1{2/(s−1) + 5s/(s^2 + 25)}
2e^t + 5cos(5t)
y′ + 2y = 4, y(0)= 1
y(t)= 2 − e^(−2t)
3/{(s+1)(s+2)}
1/ (s+1) + 1/(s+2)
L{u(t−2)(t−2)^2}
(2e^−2s) / s^3