SME
Differential Equations · Problem type
The characteristic equation
Use this for “Find the general solution of y″ − 5y′ + 6y = 0.”
Chart
Vertex (2.5, −0.25), axis of symmetry r = 2.5
Zeros: r = 2, r = 3
This is an example. Type your own values to start a new problem.
a
y″
+
(
b
)
y′
+
c
y
=
0
a
r²
+
(
b
)
r
+
c
=
0
Put y = eʳᵗ into the first line and divide by eʳᵗ: the second line, the graph’s p(r). Its roots give the solution.
Δ
Discriminant b² − 4ac
worked out (not typed)
roots
Kind of roots
worked out (not typed)
r₁
Smaller real root
worked out (not typed)
r₂
Larger real root
worked out (not typed)
α
Middle of the roots
worked out (not typed)
β
Offset of the roots from α
worked out (not typed)
Formulas
Δ
=
b
2
−
4
×
a
×
c
1
=
(−5)
2
−
4
×
1
×
6
roots
=
two
real
roots
if
Δ
>
0,
one
double
root
if
Δ
=
0,
else
complex
roots
1 > 0, so two real roots
r₁
=
−
b
− √
Δ
2 ×
a
2
=
−(−5) −
√
1
2 × 1
r₂
=
−
b
+ √
Δ
2 ×
a
3
=
−(−5) +
√
1
2 × 1
α
=
−
b
2 ×
a
2.5
=
−
−5
2 × 1
β
=
(
1
2 ×
a
)
×
√
|
Δ
|
0.5
=
(
1
2 × 1
)
×
√
|1|
Assumptions
•
Try y = ert: then y′ = rert and y″ = r²ert, and since ert is never 0, ar² + br + c = 0.
•
Two real roots give y = C₁er₁t + C₂er₂t; a double root r gives y = (C₁ + C₂t)ert.
•
Complex roots α ± βi give y = eαt(C₁ cos βt + C₂ sin βt). Here a > 0; multiply through by −1 if not.
Step-by-step
WE KNOW
a = 1, b = −5, c = 6
FIND
discriminant b² − 4ac (Δ), kind of roots (roots), smaller real root (r₁), larger real root (r₂), middle of the roots (α), offset of the roots from α (β)
Step 1 · Find discriminant b² − 4ac (Δ)
Use Δ = b² − 4 × a × c
Put
y
=
e
rt
into
the
equation
and
divide
by
e
rt
:
the
characteristic
equation.
Its
discriminant
tells
which
kind
of
roots
it
has.
r
2
−
5r
+
6
=
0
Δ
=
(−5)
2
−
4
×
1
×
6
=
1
Δ
=
1
Step 2 · Find kind of roots (roots)
Use roots = two real roots if Δ > 0, one double root if Δ = 0, else complex roots
Δ
>
0:
√
Δ
is
real,
so
the
±
gives
two
different
real
roots.
1 > 0: two real roots
roots = two real roots if 1 > 0, one double root if 1 = 0, else complex roots
roots = two real roots
Step 3 · Find smaller real root (r₁)
Use r₁ = (−b − √Δ) ÷ (2 × a)
The
quadratic
formula
with
−:
with
a
>
0
this
is
the
smaller
root.
r₁
=
−(−5) −
√
1
2 × 1
r₁
=
5 −
√
1
2
r₁
=
5 − 1
2
r₁
=
4
2
r₁
=
2
Step 4 · Find larger real root (r₂)
Use r₂ = (−b + √Δ) ÷ (2 × a)
The
quadratic
formula
with
+:
the
larger
root.
Each
real
root
r
gives
one
solution
e
rt
.
r₂
=
−(−5) +
√
1
2 × 1
y
=
C₁
e
2t
+
C₂
e
3t
r₂
=
3
Step 5 · Find middle of the roots (α)
Use α = −b ÷ (2 × a)
The
quadratic
formula
r
=
−
b
± √
Δ
2
a
splits
into
−
b
2
a
,
halfway
between
the
roots,
and
the
±
part.
α
=
−
−5
2 × 1
α
=
5
2
α
=
2.5
Step 6 · Find offset of the roots from α (β)
Use β = (1 ÷ (2 × a)) × √(|Δ|)
With
Δ
>
0
the
roots
are
r₁
=
α
−
β
and
r₂
=
α
+
β
,
and
each
root
r
gives
one
solution
e
rt
.
β
=
(
1
2 ×
a
)
×
√
Δ
β
=
(
1
2 × 1
)
×
√
1
r
=
5 ±
√
1
2
=
2
or
3
β
=
0.5
Check
Put the numbers back in. Both sides should match.
1
=
(−5)
2
−
4
×
1
×
6
✓
1 > 0, so two real roots ✓
2
=
−(−5) −
√
1
2 × 1
✓
3
=
−(−5) +
√
1
2 × 1
✓
2.5
=
−
−5
2 × 1
✓
0.5
=
(
1
2 × 1
)
×
√
|1|
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.