SME
Differential Equations · Problem type
Inverse transform by partial fractions
Use this for “Find the inverse Laplace transform of (s + 3) ÷ ((s + 1)(s + 2)), and its value at t = 1.”
Chart
y(1) ≈ 0.6004
This is an example. Type your own values to start a new problem.
s
+
(s
−
(
)
)(s
−
(
)
)
=
A
s
−
(
)
+
B
s
−
(
)
y(
t
)
=
y(t)
Formulas
A
=
p
×
r₁
+
q
r₁
−
r₂
2
=
1 × (−1) + 3
−1 − (−2)
B
=
p
×
r₂
+
q
r₂
−
r₁
−1
=
1 × (−2) + 3
−2 − (−1)
y(
t
)
=
A
×
e
r₁
×
t
+
B
×
e
r₂
×
t
0.6004
=
2
×
e
−1 × 1
+
(−1)
×
e
−2 × 1
Assumptions
•
The bottom has two different real roots r₁ and r₂, and the top is of lower degree, so Y = A ÷ (s − r₁) + B ÷ (s − r₂).
•
Cover-up rule: multiply by (s − r₁) and put s = r₁ to get A = (pr₁ + q) ÷ (r₁ − r₂); B the same way.
•
From the table, the inverse of 1 ÷ (s − r) is ert, so y(t) = Aer₁t + Ber₂t.
Step-by-step
WE KNOW
p = 1, q = 3, r₁ = −1, r₂ = −2, t = 1
FIND
numerator over s − r₁ (A), numerator over s − r₂ (B), inverse transform at t (y(t))
Step 1 · Find numerator over s − r₁ (A)
Use A = (p × r₁ + q) ÷ (r₁ − r₂)
Cover
up
the
factor
(s
−
r₁
)
and
put
s
=
r₁
into
what
is
left:
the
top
ps
+
q
over
s
−
r₂
.
A
=
1 × (−1) + 3
−1 − (−2)
A
=
2
Step 2 · Find numerator over s − r₂ (B)
Use B = (p × r₂ + q) ÷ (r₂ − r₁)
Cover
up
(s
−
r₂
)
and
put
s
=
r₂
.
The
two
pieces
add
back
up
to
Y(s).
B
=
1 × (−2) + 3
−2 − (−1)
B
=
1
−1
=
−1
Y(s)
=
s + 3
(s + 1)(s + 2)
=
2
s + 1
−
1
s + 2
B
=
−1
Step 3 · Find inverse transform at t (y(t))
Use y(t) = A × er₁ × t + B × er₂ × t
Read
each
piece
back
from
the
table:
A
s −
r₁
comes
from
A
e
r₁t
,
and
B
s −
r₂
from
B
e
r₂t
.
y(
t
)
=
2
e
−
t
−
e
−2
t
y(1)
=
2e⁻¹
−
e
-2
=
0.735759
−
0.135335
=
0.6004
y(
t
)
=
0.6004
Check
Put the numbers back in. Both sides should match.
2
=
1 × (−1) + 3
−1 − (−2)
✓
−1
=
1 × (−2) + 3
−2 − (−1)
✓
0.6004
=
2
×
e
−1 × 1
+
(−1)
×
e
−2 × 1
✓
Topic check
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