SME
Differential Equations
Laplace transforms
Use this for “Solve y′ + 0.5y = 2u(t − 2), y(0) = 1, by the Laplace transform. What is y(4)?”
Chart
y(4) ≈ 2.664
This is an example. Type your own values to start a new problem.
y′
+
k
y
=
A
u(t
−
τ
)
y(0)
=
y₀
y(
t
)
=
y(t)
Formulas
y(
t
)
=
y₀
×
e
−
k
×
t
+
A
k
×
(1
−
e
−
k
× (
t
−
τ
)
)
2.6638
=
1
×
e
−0.5 × 4
+
2
0.5
×
(1
−
e
−0.5 × (4 − 2)
)
Assumptions
•
The transform turns the equation into algebra: L{y′} = sY − y₀ and L{u(t − τ)} = e−τs ÷ s.
•
A factor e−τs shifts by τ: it turns f(t) back into f(t − τ)u(t − τ), which is 0 before τ.
•
Partial fractions split A ÷ (s(s + k)) into (A ÷ k)(1 ÷ s − 1 ÷ (s + k)), whose inverse is (A ÷ k)(1 − e−kt).
Step-by-step
WE KNOW
k = 0.5, A = 2, τ = 2, y₀ = 1, t = 4
FIND
solution at t (y(t))
Step 1 · Find solution at t (y(t))
Use y(t) = y₀ × e−k × t + A ÷ k × (1 − e−k × (t − τ))
Transform,
solve
for
Y(s)
and
split
it.
Each
e
−τs
term
comes
back
shifted
by
τ
,
and
for
t
≥
τ
the
step
is
on:
u(
t
−
τ
)
=
1.
4 ≥ 2: after the step
y′
+
0.5y
=
2u(
t
−
2),
y(0)
=
1
sY
−
1
+
0.5Y
=
(
2
s
)
e
−2s
Y(s)
=
1
s + 0.5
+
(
2
s(s + 0.5)
)
e
−2s
=
1
s + 0.5
+
4(
1
s
−
1
s + 0.5
)
e
−2s
y(
t
)
=
e
−0.5
t
+
4(1
−
e
−0.5(
t
− 2)
)u(
t
−
2)
y(4)
=
e
-2
+
4(1
−
e
-1
)
=
0.135335
+
2.52848
=
2.6638
y(
t
)
=
2.6638
Check
Put the numbers back in. Both sides should match.
2.6638
=
1
×
e
−0.5 × 4
+
2
0.5
×
(1
−
e
−0.5 × (4 − 2)
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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Inverse transform by partial fractions
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The steps of solving by Laplace transform
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