Case 2 - repeated real root — reduction of order: Question bank

Recall the anchor facts before you start:

A repeated root $r$ occurs when the discriminant $b^{2} - 4 a c = 0$ , giving one root $r = - \frac{b}{2 a}$ .
The two independent solutions are $e^{r x}$ and $x e^{r x}$ , so $y = (C_{1} + C_{2} x) e^{r x}$ .
Independence is certified by the Wronskian $W = e^{2 r x} \neq = 0$ .

Spot the error

r^{2} - 6 r + 9 = 0

has roots

3

and

- 3

, so

y = C_{1} e^{3 x} + C_{2} e^{- 3 x}

Error:

r^{2} - 6 r + 9 = (r - 3)^{2}

, a double root

r = 3

, not

\pm 3

; the correct solution is

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

"The reduction gives

v = C_{1} + C_{2} x

, so

y = C_{1} + C_{2} x e^{r x}

Error: the whole

v

multiplies

y_{1} = e^{r x}

, so

y = (C_{1} + C_{2} x) e^{r x}

; the

C_{1}

term must also carry the

e^{r x}

factor.

"Since

v^{''} = 0

means

v

is constant, the second solution is just another

e^{r x}

Error:

v^{''} = 0

gives

v = C_{1} + C_{2} x

, a linear function, not a constant; the new piece is the

C_{2} x

term producing

x e^{r x}

"For

4 y^{''} + 12 y^{'} + 9 y = 0

the repeated root is

r = - b /2 a = - 12/2 = - 6

Error:

a = 4

, so

r = - b /2 a = - 12/8 = - 3/2

; the student forgot to divide by

a

, not just by

2

"Because both roots equal

r

, the Wronskian of the two exponentials is zero, so no independent pair exists."

Error: there is only one exponential

e^{r x}

; the correct partner is

x e^{r x}

, and the Wronskian of

e^{r x}

with

x e^{r x}

e^{2 r x} \neq = 0

"Applying

y (0) = 1

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

gives

C_{1} + C_{2} = 1

Error: at

x = 0

the

C_{2} x

term vanishes and

e^{0} = 1

, so

y (0) = C_{1} = 1

;

C_{2}

is fixed later by

y^{'} (0)

"The characteristic equation of

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

factors as

(r + 1) (r - 1)

, giving distinct roots."

Error: it factors as

(r + 1)^{2}

, a repeated root

r = - 1

; the student mis-signed the constant term.

Why questions

Why does a second-order ODE require exactly two independent solutions?

Its solution set is a 2-dimensional vector space, so any solution is a unique combination of two basis functions — enough freedom to match two initial data

y (x_{0}), y^{'} (x_{0})

Why do BOTH the

v

and

v^{'}

terms disappear in the repeated-root reduction?

The

v

-coefficient is

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

because

r

is a root, and the

v^{'}

-coefficient is

2 a r + b = 0

because

r = - b /2 a

; the double root makes both hold simultaneously.

Why can't we just pick

y_{2} = e^{r_{2} x}

with a "second" root as in Case 1?

There is no second distinct root — the discriminant is zero so both roots collapse to the same

r

, leaving Case 1's recipe with only one function.

Why does sticking an

x

onto

e^{r x}

(rather than any other factor) produce the missing solution?

Reduction of order forces

v^{''} = 0

, whose non-constant solution is the linear

v = x

; the algebra of the double root singles out

x

, and the coalescing-roots limit confirms it as

\partial_{r} e^{r x}

Why does the characteristic equation method fail to hand us two solutions here, unlike in the distinct-root case?

The trial

y = e^{r x}

turns the ODE into a quadratic; a double root is one number, so it yields one exponential — the geometry of "two coincident roots" leaves the algebra one solution short.

Why is the sign of

r

irrelevant to whether

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

are independent?

Independence depends on the Wronskian

e^{2 r x}

being nonzero, and exponentials of any real exponent are strictly positive, so no choice of

r

can make them dependent.

Why does the general solution automatically satisfy any initial-value problem for this ODE?

The two free constants

C_{1}, C_{2}

give exactly two degrees of freedom, matching the two conditions

y (x_{0}), y^{'} (x_{0})

; the nonzero Wronskian guarantees the resulting system is solvable.

Edge cases

What are the solutions when the repeated root is

r = 0

(e.g.

y^{''} = 0

With

r = 0

e^{r x} = 1

and

x e^{r x} = x

, so

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

— the straight lines, still two independent functions.

a \to 0

so the equation degenerates to

b y^{'} + cy = 0

, does the repeated-root picture still apply?

No — the ODE drops to first order, has a 1-dimensional solution space, and only one exponential; the whole "missing partner" story needs a genuine second-order term.

For a triple root

r

in a third-order ODE, what is the solution basis?

{e^{r x}, x e^{r x}, x^{2} e^{r x}}

— Higher-order repeated roots: multiplicity

m

supplies powers

x^{0}

through

x^{m - 1}

times

e^{r x}

As two nearly-equal roots

r

and

r + ε

merge, what happens to the pair of exponential solutions?

They become nearly parallel; their normalized difference

\frac{e ^{(r + ε) x} - e ^{r x}}{ε}

approaches

x e^{r x}

, so the basis smoothly deforms from two exponentials to

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

What is the long-run behaviour of

y = (C_{1} + C_{2} x) e^{r x}

when

r < 0

It tends to

0

, because the exponential decay

e^{r x}

eventually beats the linear growth of

C_{2} x

— exponential wins over polynomial.

What is the long-run behaviour when

r = 0

but

C_{2} \neq = 0

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

grows (or falls) linearly without bound, since there is no decaying exponential to tame the

x

term.

Can

e^{r x}

and

x e^{r x}

ever be simultaneously zero at some finite

x

No —

e^{r x} > 0

always, so

e^{r x}

is never zero and

x e^{r x}

vanishes only at

x = 0

, where the two are still independent by the Wronskian.

Recall One-line summary of every trap

A double root gives one exponential; the honest fix is $x e^{r x}$ , forced by $v^{''} = 0$ and confirmed by the nonzero Wronskian $e^{2 r x}$ . Never collapse the constants, never drop the $x$ , always divide by $a$ in $r = - b /2 a$ , and remember decay needs $r < 0$ .

Connections

Case 2 - repeated real root — reduction of order (index 4.6.12) — parent derivation these traps guard.
Case 1 distinct real roots — the contrast most errors confuse it with.
Case 3 complex roots — the other sign of the discriminant.
Reduction of order (general method) — why $v^{''} = 0$ appears.
Wronskian and linear independence — the independence certificate.
Higher-order repeated roots — the triple-root edge case.
Characteristic equation of linear ODEs — origin of the root $r$ .

Question bank — Case 2 - repeated real root — reduction of order

True or false — justify

Spot the error

Why questions

Edge cases

Connections