Homogeneous with constant coefficients — characteristic equation: Question bank

Quick symbol reminder so nothing here is used unexplained:

$y^{'} = \frac{d y}{d x}$ means "how fast $y$ changes as $x$ moves"; $y^{''}$ is how fast that rate changes.
$r$ is the number in the guess $y = e^{r x}$ — the root of the characteristic equation $a r^{2} + b r + c = 0$ .
$Δ = b^{2} - 4 a c$ is the discriminant: the thing under the square root that decides which of the three cases (Different / Double / Dizzy) you are in.
$α \pm i β$ means a complex root with real part $α$ (growth/decay) and imaginary part $β$ (wiggle speed).

True or false — justify

TF1. "Every second-order homogeneous constant-coefficient ODE has exactly two independent solutions."

True — the order is 2, and the linearity of the operator guarantees a 2-dimensional solution space; even a repeated root still yields two independent pieces

e^{r x}

and

x e^{r x}

TF2. "If

y_{1}

and

y_{2}

both solve

L [y] = 0

, then

3 y_{1} - 7 y_{2}

also solves it."

True — the operator

L

is linear, so

L [3 y_{1} - 7 y_{2}] = 3 L [y_{1}] - 7 L [y_{2}] = 0

; this superposition only works because the equation is homogeneous (right side

0

). See Linear Differential Operators and Superposition.

TF3. "You can always divide the substituted equation by

e^{r x}

to get the polynomial."

True —

e^{r x}

is never zero for any real or complex

x

, so dividing is legal and loses no solutions; this is the step that turns calculus into algebra.

TF4. "For

a y^{''} + b y^{'} + cy = 0

the characteristic equation is

r^{2} + b r + c = 0

False — it is

a r^{2} + b r + c = 0

; dropping the leading

a

changes both the roots and the discriminant

Δ = b^{2} - 4 a c

unless

a = 1

TF5. "Complex roots mean the ODE has no real solutions."

False — the ODE still has purely real solutions

e^{α x} cos β x

and

e^{α x} sin β x

; the complex roots are just an intermediate bookkeeping device, converted back to real form via Euler. See Euler's Formula and Complex Exponentials.

TF6. "A repeated root always corresponds to

Δ = 0

True — a double root exists exactly when the square root in the quadratic formula vanishes, i.e.

b^{2} - 4 a c = 0

, giving the single value

r = - b / (2 a)

TF7. "If

α < 0

the solution decays; if

α > 0

it blows up."

True — the envelope is

e^{α x}

, which shrinks when

α < 0

and grows when

α > 0

; this is exactly the growth/decay knob of a Damped Harmonic Oscillator.

TF8. "

e^{2 x}

and

e^{3 x}

could be dependent if the constants line up right."

False — distinct-exponent exponentials are always linearly independent; their Wronskian is

(3 - 2) e^{5 x} \neq = 0

for all

x

. See Wronskian and Linear Independence of Solutions.

TF9. "The value of

β

(imaginary part) sets how fast the solution oscillates."

True —

β

is the angular frequency inside

cos β x

and

sin β x

; larger

β

means more wiggles per unit

x

, while

α

leaves the frequency untouched.

Spot the error

SE1. "Repeated root

r = 3

, so

y = c_{1} e^{3 x} + c_{2} e^{3 x}

The two terms collapse into

(c_{1} + c_{2}) e^{3 x}

— a single constant, so this is only one solution; the correct second solution is

x e^{3 x}

, giving

y = (c_{1} + c_{2} x) e^{3 x}

SE2. "Roots

r = - 1 \pm 2 i

, so the answer is

y = c_{1} e^{(- 1 + 2 i) x} + c_{2} e^{(- 1 - 2 i) x}

and we stop."

Not wrong, but incomplete for a real problem: convert with Euler to the real form

y = e^{- x} (c_{1} cos 2 x + c_{2} sin 2 x)

so the solution is real-valued.

SE3. "For

y^{''} + 4 y = 0

, char. eq.

r^{2} + 4 = 0

gives

r = \pm 4

Error:

r^{2} = - 4

, not

16

, so

r = \pm 2 i

, not

\pm 4

; the roots are imaginary, giving oscillation

c_{1} cos 2 x + c_{2} sin 2 x

SE4. "For

2 y^{''} + 3 y^{'} + y = 0

the discriminant is

Δ = 3^{2} - 4 (1) (1) = 5

Error:

a = 2

c = 1

, so

Δ = b^{2} - 4 a c = 9 - 8 = 1

, not

5

; forgetting the leading coefficient

a

is the classic slip.

SE5. "

y = e^{- x} sin 2 x

; differentiating gives

y^{'} = e^{- x} \cdot 2 cos 2 x

Error: this is a product, so the product rule is mandatory —

y^{'} = - e^{- x} sin 2 x + 2 e^{- x} cos 2 x

; dropping the first term loses the decay's contribution.

SE6. "Since

r = 0

is a root of

r^{2} - 3 r = 0

, the solution

e^{0 \cdot x}

is trivial and can be ignored."

Error:

e^{0 x} = 1

is a genuine, non-trivial constant solution; the full answer is

y = c_{1} + c_{2} e^{3 x}

, and dropping the constant loses an entire dimension.

SE7. "The characteristic equation of

y^{'''} - y = 0

r^{2} - 1 = 0

Error: the order is 3, so it must be

r^{3} - 1 = 0

; the polynomial degree always equals the ODE order.

Why questions

WHY1. Why do we guess an exponential rather than, say, a polynomial or a sine?

Because differentiation must return a scaled copy of the same function for the linear combination to cancel to zero, and

e^{r x}

is the unique family with

\frac{d}{d x} e^{r x} = r e^{r x}

; sines/cosines appear only as the disguise exponentials wear when

r

is complex.

WHY2. Why does a double root force an extra factor of

x

rather than

x^{2}

ln x

Factoring the operator as

(D - r_{0})^{2}

and solving the resulting first-order chain leaves exactly one constant to integrate, which produces a linear term

c_{2} x

; a triple root would then need

x^{2}

. See Second Order Linear ODE — Reduction of Order.

WHY3. Why must the ODE be homogeneous for superposition to work?

Superposition needs

L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}] = 0

, which requires both pieces to hit

0

; with a nonzero right side the combination would sum the forcing terms instead of cancelling. See Method of Undetermined Coefficients.

WHY4. Why does the real part

α

of a complex root not affect the oscillation frequency?

Because

α

lives in the envelope

e^{α x}

(a pure amplitude scaling) while the frequency lives entirely in

β

inside

cos β x

; multiplying a wave by a shrinking amplitude changes its size, not how often it crosses zero.

WHY5. Why can we take real and imaginary parts of a complex solution and get two independent real solutions?

Because the ODE has real coefficients, so if

u + i v

solves it then

u

and

v

solve it separately;

cos β x

and

sin β x

are not scalar multiples of each other, so they are independent (nonzero Wronskian).

WHY6. Why does the discriminant

Δ

, and not the individual coefficients, decide the shape of the solution?

Δ = b^{2} - 4 a c

is precisely the quantity under the square root in

r = \frac{- b \pm Δ}{2 a}

, so its sign alone determines whether the roots are two reals, one real, or a complex pair — that is what fixes the case.

WHY7. Why is dividing by

e^{r x}

safer than dividing by, say,

y

itself?

e^{r x}

is provably nonzero everywhere, so no solutions are lost; dividing by

y

could destroy the point where

y = 0

and introduce spurious restrictions.

Edge cases

EC1. "What if one root is

r = 0

Then

e^{0 x} = 1

is a constant solution; for

r = 0

and

r = k

the answer is

y = c_{1} + c_{2} e^{k x}

, and a double root at

0

gives

y = c_{1} + c_{2} x

EC2. "What if both roots are

r = 0

(double root at zero)?"

The equation is essentially

y^{''} = 0

, so

y = c_{1} + c_{2} x

— the

x e^{r x}

rule with

r = 0

collapses to a straight line, matching plain double integration.

EC3. "What if the roots are complex with

α = 0

, like

\pm i β

The envelope

e^{0 x} = 1

disappears, leaving undamped pure oscillation

y = c_{1} cos β x + c_{2} sin β x

— the ideal frictionless spring.

EC4. "What if

a = 0

, so there is no

y^{''}

term?"

Then it is not second-order at all; it drops to a first-order equation

b y^{'} + cy = 0

with a single root

r = - c / b

and solution

y = c_{1} e^{r x}

— the whole three-case machinery does not apply.

EC5. "What if

b = 0

and

c > 0

, e.g.

y^{''} + cy = 0

Then

Δ = - 4 a c < 0

automatically, giving pure imaginary roots and undamped oscillation;

b

is the damping term, so removing it removes all decay.

EC6. "What if two exponents are very close but distinct, like

r = 3

and

r = 3.0001

They stay Case 1 with two separate exponentials; as the gap shrinks toward zero the pair

e^{r_{1} x}, e^{r_{2} x}

continuously deforms into

e^{r x}, x e^{r x}

, which is why the repeated case borrows the

x

factor — it is the limit of the distinct case.

EC7. "What if the initial conditions are applied to

e^{α x} (c_{1} cos β x + c_{2} sin β x)

x = 0

x = 0

the envelope is

e^{0} = 1

cos 0 = 1

sin 0 = 0

, so

y (0) = c_{1}

immediately; but

y^{'} (0)

still needs the full product rule across both the envelope and the trig factor before substituting.

Connections

Linear Differential Operators and Superposition
Euler's Formula and Complex Exponentials
Wronskian and Linear Independence of Solutions
Method of Undetermined Coefficients
Damped Harmonic Oscillator
Second Order Linear ODE — Reduction of Order