SME
Differential Equations · Problem type
Transform of c·tⁿeᵃᵗ
Use this for “Find the Laplace transform of 3t²e²ᵗ, and its value at s = 4.”
Table
Point s (s)
Transform at the point s (F(s))
This is an example. Type your own values to start a new problem.
f(t)
=
c
t
e
t
F(
s
)
=
F(s)
Formulas
F(
s
)
=
c
×
n
!
÷
(
s
−
a
)
n
+ 1
0.75
=
3
×
2!
÷
(4 − 2)
2 + 1
Assumptions
•
L{f} = F(s) = ∫ from 0 to ∞ of e−stf(t) dt, which exists for c·tⁿeat only when s > a.
•
From the table, L{tⁿ} = n! ÷ sⁿ⁺¹; the factor eat shifts s to s − a, and a constant c stays in front.
•
n is a whole number from 0 to 6 (n = 0 is c·eat, whose transform is c ÷ (s − a)).
Step-by-step
WE KNOW
c = 3, n = 2, a = 2, s = 4
FIND
transform at the point s (F(s))
Step 1 · Find transform at the point s (F(s))
Use F(s) = c × n! ÷ (s − a)ⁿ⁺¹
Read
L{tⁿ}
=
n
!
sⁿ
⁺¹
from
the
table,
put
s
−
a
for
s
because
of
e
at
,
and
keep
c
in
front.
L{3t²
e
2t
}
=
3
×
2!
÷
(
s
− 2)
3
=
6
÷
(
s
− 2)
3
F(4)
=
6
÷
(4 − 2)
3
=
6
8
=
0.75
F(
s
)
=
0.75
Check
Put the numbers back in. Both sides should match.
0.75
=
3
×
2!
÷
(4 − 2)
2 + 1
✓
Topic check
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