SME
Differential Equations · Problem type
Solve x′ = Ax from a starting point
Use this for “Solve x′ = x + 2y, y′ = 2x + y with x(0) = 3, y(0) = 1. Where is the solution at t = 0.5?”
Diagram
Trace T = 2, determinant D = −3, T² − 4D = 16
λ = 3 along ⟨1, 1⟩ and λ = −1 along ⟨1, −1⟩ (the dashed lines)
The origin is a saddle
The path runs from (3, 1) at t = 0 to (9.57, 8.357) at t = 0.5
This is an example. Type your own values to start a new problem.
x′
=
a
x
+
b
y,
y′
=
c
x
+
d
y
x(0)
=
x₀
,
y(0)
=
y₀
At
t
=
:
x
=
x(t)
,
y
=
y(t)
T
Trace of A
worked out (not typed)
D
Determinant of A
worked out (not typed)
Δ
Discriminant T² − 4D
worked out (not typed)
λ₁
Larger eigenvalue
worked out (not typed)
λ₂
Smaller eigenvalue
worked out (not typed)
m₁
Slope of eigenvector v₁
worked out (not typed)
m₂
Slope of eigenvector v₂
worked out (not typed)
c₁
Weight of v₁
worked out (not typed)
c₂
Weight of v₂
worked out (not typed)
Formulas
T
=
a
+
d
2 = 1 + 1
D
=
a
×
d
−
b
×
c
−3 = 1 × 1 − 2 × 2
Δ
=
T
2
−
4
×
D
16
=
2
2
−
4
×
(−3)
λ₁
=
T
+ √
Δ
2
3
=
2 +
√
16
2
λ₂
=
T
− √
Δ
2
−1
=
2 −
√
16
2
m₁
=
λ₁
−
a
b
1
=
3 − 1
2
m₂
=
λ₂
−
a
b
−1
=
−1 − 1
2
c₁
=
y₀
−
m₂
×
x₀
m₁
−
m₂
2
=
1 − (−1) × 3
1 − (−1)
c₂
=
x₀
−
c₁
1 = 3 − 2
x(
t
)
=
c₁
×
e
λ₁
×
t
+
c₂
×
e
λ₂
×
t
9.5699
=
2
×
e
3 × 0.5
+
1
×
e
−1 × 0.5
y(
t
)
=
c₁
×
m₁
×
e
λ₁
×
t
+
c₂
×
m₂
×
e
λ₂
×
t
8.3568
=
2
×
1
×
e
3 × 0.5
+
1
×
(−1)
×
e
−1 × 0.5
Assumptions
•
With two real, different eigenvalues, x(t) = c₁eλ₁tv₁ + c₂eλ₂tv₂: each eigenvector part grows or shrinks on its own.
•
Each eigenvector is written ⟨1, m⟩, with m = (λ − a) ÷ b from the first row of (A − λI)v = 0 (b ≠ 0).
•
c₁ and c₂ make the start match: c₁ + c₂ = x(0) and c₁m₁ + c₂m₂ = y(0).
Step-by-step
WE KNOW
a = 1, b = 2, c = 2, d = 1, x₀ = 3, y₀ = 1, t = 0.5
FIND
trace of A (T), determinant of A (D), discriminant T² − 4D (Δ), larger eigenvalue (λ₁), smaller eigenvalue (λ₂), slope of eigenvector v₁ (m₁), slope of eigenvector v₂ (m₂), weight of v₁ (c₁), weight of v₂ (c₂), x at time t (x(t)), y at time t (y(t))
Step 1 · Find trace of A (T)
Use T = a + d
The
trace
adds
the
diagonal
of
A,
and
it
equals
λ₁
+
λ₂
,
the
sum
of
the
eigenvalues.
T
=
1
+
1
T
=
2
Step 2 · Find determinant of A (D)
Use D = a × d − b × c
A
=
[[1,
2],
[2,
1]].
Its
determinant
equals
λ₁
λ₂
,
the
product
of
the
eigenvalues:
negative
when
they
have
opposite
signs.
D
=
1
×
1
−
2
×
2
D
=
1
−
4
D
=
−3
Step 3 · Find discriminant T² − 4D (Δ)
Use Δ = T² − 4 × D
det(A
−
λI)
=
0
is
λ
2
−
Tλ
+
D
=
0.
Its
discriminant
says
whether
the
eigenvalues
are
real
(
Δ
≥
0)
or
complex
(
Δ
<
0).
Δ
=
2
2
−
4
×
(−3)
Δ
=
4
−
4
×
(−3)
Δ
=
4
−
(−12)
Δ
=
16
Step 4 · Find larger eigenvalue (λ₁)
Use λ₁ = (T + √Δ) ÷ 2
The
larger
eigenvalue:
a
solution
starting
on
its
eigenvector
line
stays
on
it,
scaled
by
e
λ₁t
.
λ₁
=
2 +
√
16
2
λ₁
=
2 + 4
2
λ₁
=
6
2
λ₁
=
3
Step 5 · Find smaller eigenvalue (λ₂)
Use λ₂ = (T − √Δ) ÷ 2
The smaller eigenvalue: its eigenvector line is the second straight-line direction.
λ₂
=
2 −
√
16
2
λ₂
=
2 − 4
2
λ₂
=
−
2
2
λ₂
=
−1
Step 6 · Find slope of eigenvector v₁ (m₁)
Use m₁ = (λ₁ − a) ÷ b
The
first
row
of
(A
−
λ₁I)v
=
0
reads
(
a
−
λ₁
)
×
1
+
b
×
m₁
=
0,
so
v₁
=
⟨1,
m₁
⟩.
m₁
=
3 − 1
2
m₁
=
2
2
m₁
=
1
Step 7 · Find slope of eigenvector v₂ (m₂)
Use m₂ = (λ₂ − a) ÷ b
The
same
row
with
λ₂
gives
v₂
=
⟨1,
m₂
⟩.
m₂
=
−1 − 1
2
m₂
=
−
2
2
m₂
=
−1
Step 8 · Find weight of v₁ (c₁)
Use c₁ = (y₀ − m₂ × x₀) ÷ (m₁ − m₂)
At
t
=
0
both
exponentials
are
1:
c₁
+
c₂
=
x₀
and
c₁
m₁
+
c₂
m₂
=
y₀
.
Take
m₂
times
the
first
from
the
second.
c₁
=
1 − (−1) × 3
1 − (−1)
c₁
=
1 − (−3)
2
c₁
=
4
2
c₁
=
2
Step 9 · Find weight of v₂ (c₂)
Use c₂ = x₀ − c₁
Then
the
first
equation
gives
c₂
.
c₂
=
3
−
2
c₂
=
1
Step 10 · Find x at time t (x(t))
Use x(t) = c₁ × eλ₁ × t + c₂ × eλ₂ × t
The
x
parts
of
v₁
and
v₂
are
both
1,
so
x(
t
)
adds
the
two
weights,
each
scaled
by
its
own
e
λt
.
x(
t
)
=
2
×
e
3 × 0.5
+
1
×
e
−1 × 0.5
x(0.5) = 2 × 4.48169 + 1 × 0.606531 = 9.5699
x(
t
)
=
9.5699
Step 11 · Find y at time t (y(t))
Use y(t) = c₁ × m₁ × eλ₁ × t + c₂ × m₂ × eλ₂ × t
The
y
parts
carry
the
slopes
m₁
and
m₂
.
y(
t
)
=
2
×
1
×
e
3 × 0.5
+
1
×
(−1)
×
e
−1 × 0.5
y(0.5) = 2 × 4.48169 − 1 × 0.606531 = 8.3568
y(
t
)
=
8.3568
Check
Put the numbers back in. Both sides should match.
2 = 1 + 1 ✓
−3 = 1 × 1 − 2 × 2 ✓
16
=
2
2
−
4
×
(−3)
✓
3
=
2 +
√
16
2
✓
−1
=
2 −
√
16
2
✓
1
=
3 − 1
2
✓
−1
=
−1 − 1
2
✓
2
=
1 − (−1) × 3
1 − (−1)
✓
1 = 3 − 2 ✓
9.5699
=
2
×
e
3 × 0.5
+
1
×
e
−1 × 0.5
✓
8.3568
=
2
×
1
×
e
3 × 0.5
+
1
×
(−1)
×
e
−1 × 0.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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