Cauchy-Euler (Equidimensional) equation: Question bank

Recall the object we are reasoning about: the equation $a x^{2} y^{''} + b x y^{'} + c y = 0$ , its indicial equation $a m (m - 1) + b m + c = 0$ (equivalently $a m^{2} + (b - a) m + c = 0$ ), and the guess $y = x^{m}$ . Here $y^{''}$ means "second derivative of $y$ ", $ln x$ is the natural logarithm, and $C_{1}, C_{2}$ are arbitrary constants.

True or false — justify

Every second-order Cauchy-Euler equation has at least one solution of the pure form

y = x^{m}

True — the indicial quadratic

a m^{2} + (b - a) m + c = 0

always has at least one root

m

(real or complex), and

x^{m}

built from that root always solves the equation; only the second solution may need a

ln x

or a

cos / sin

dressing.

The indicial equation of

a x^{2} y^{''} + b x y^{'} + cy = 0

a m^{2} + bm + c = 0

False — differentiating

x^{m}

twice gives

m (m - 1)

, so

x^{2} y^{''}

contributes

am (m - 1) = a m^{2} - am

; the correct equation is

a m^{2} + (b - a) m + c = 0

, carrying an extra

- am

If the indicial roots are

m_{1} = 2, m_{2} = 2

(repeated), the general solution is

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

False — the repeated-root partner is a factor of

ln x

, not

x

; the answer is

(C_{1} + C_{2} ln x) x^{2}

. The "extra

x

" idea is borrowed illegally from constant-coefficient ODEs.

x^{i}

(with

i = - 1

) is a genuine complex-valued solution when the roots are

m = \pm i

True —

x^{i} = e^{i l n x} = cos (ln x) + i sin (ln x)

by Euler's Formula; it solves the ODE, and its real and imaginary parts give the two real solutions.

Because the equation is "equidimensional", scaling

x \to λ x

leaves the solution set unchanged.

True — every term

x^{k} y^{(k)}

has the same dimension, so the ODE is scale-invariant; if

y (x)

solves it, so does

y (λ x)

, which is why power functions (themselves scale-covariant) are the natural solutions.

A Cauchy-Euler equation is a constant-coefficient ODE in disguise.

True — the substitution

t = ln x

turns it into

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

with constant coefficients; the "disguise" is the variable

x

, and

t = ln x

removes it.

The origin

x = 0

is an ordinary point of

x^{2} y^{''} + b x y^{'} + cy = 0

False — at

x = 0

the leading coefficient

x^{2}

vanishes, so

x = 0

is a singular point (in fact a regular singular point, see Frobenius Method & Regular Singular Points); solutions like

x^{- 1}

ln x

blow up there.

For

x < 0

the solution

x^{m}

must be replaced by

∣ x ∣^{m}

and

ln x

ln ∣ x ∣

True —

ln x

and non-integer powers are undefined for negative

x

; on the interval

x < 0

we use

∣ x ∣

so the expressions stay real and defined.

Spot the error

x^{2} y^{''} - 2 x y^{'} - 4 y = 0

gives indicial

m^{2} - 2 m - 4 = 0

." — where is the slip?

The

x^{2} y^{''}

term gives

m (m - 1) = m^{2} - m

, not

m^{2}

; the correct indicial is

m^{2} - m - 2 m - 4 = m^{2} - 3 m - 4 = 0

. The writer forgot the

- m

from

m (m - 1)

"Repeated root

m = 3

, so

y = (C_{1} + C_{2} x^{3} ln x)

." — fix it.

Both terms must carry the base power

x^{3}

; the answer is

y = (C_{1} + C_{2} ln x) x^{3}

. The

ln x

multiplies the second copy of

x^{3}

, it does not replace the first

x^{3}

or float free of it.

"Complex roots

m = α \pm i β

give

y = x^{α} (C_{1} cos β x + C_{2} sin β x)

." — what is wrong?

The oscillation is in

ln x

, not

x

: it must be

cos (β ln x)

and

sin (β ln x)

, since

x^{i β} = e^{i β l n x}

. Writing

cos β x

imports the constant-coefficient answer by mistake.

"To solve

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

I use undetermined coefficients with the guess

y = e^{r x}

." — why does this fail?

Differentiating

e^{r x}

gives another

e^{r x}

but the

x^{2}

x

coefficients then leave

x

-powers uncancelled — the terms do not collapse. The equidimensional structure demands

y = x^{m}

so that

x^{k} \frac{d ^{k}}{d x ^{k}} x^{m} \propto x^{m}

"After

t = ln x

, the equation

x^{2} y^{''} + b x y^{'} + cy = 0

becomes

a \overset{y}{¨} + b \overset{y}{˙} + cy = 0

." — correct it.

The

\overset{y}{˙}

coefficient is

b - a

, not

b

, because

x^{2} y^{''} = \overset{y}{¨} - \overset{y}{˙}

contributes an extra

- a \overset{y}{˙}

. The transformed equation is

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

Why questions

Why do we guess a power

x^{m}

here instead of an exponential

e^{r x}

(which works for Constant-Coefficient Linear ODEs)?

Because differentiating

x^{m}

lowers the power by one while the coefficient

x^{k}

raises it back by

k

, so

x^{k} y^{(k)}

stays proportional to

x^{m}

and every term collapses to (number)

\cdot x^{m}

. Exponentials do this trick only when coefficients are constant.

Why does a repeated root force a

ln x

rather than a second power?

In the

t = ln x

world the repeated root gives

e^{m t}

and

t e^{m t}

(standard Reduction of Order result); translating back with

t = ln x

turns

t

into

ln x

, giving

x^{m}

and

(ln x) x^{m}

Why is the transformed characteristic equation in

t

identical to the indicial equation in

x

Because both encode the same collapse: substituting

y = e^{m t}

into

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

gives

a m^{2} + (b - a) m + c = 0

, and

e^{m t} = e^{m l n x} = x^{m}

is exactly the power guess. The two methods are one calculation viewed in two variables.

Why does the naive quadratic-in-

m

have

(b - a)

as its middle coefficient and not

b

The second-derivative term produces

m (m - 1) = m^{2} - m

; the extra

- am

merges with

+ bm

to give

(b - a) m

. This shift is the single most common trap in the whole topic.

Why does the Euler identity produce oscillation in

ln x

, not in

x

Because

x^{i β} = e^{(i β) l n x}

, so the angle fed into

cos

sin

θ = β ln x

; the argument is whatever multiplies

i

in the exponent, and that is

ln x

Why can we solve a non-homogeneous Cauchy-Euler equation by first substituting

t = ln x

The substitution makes coefficients constant, so we may then use undetermined coefficients or Variation of Parameters in

t

freely, and only at the end translate

t = ln x

back — the equidimensional structure guarantees this conversion is exact.

Why is the equation called "equidimensional" and how does that predict power-law solutions?

Each term

x^{k} y^{(k)}

shares one physical dimension, so the equation has no built-in length scale; a scale-free equation is naturally solved by scale-covariant functions, and pure powers

x^{m}

are precisely the functions that scale cleanly.

Edge cases

If both indicial roots are zero (

m_{1} = m_{2} = 0

), what is the general solution?

y = (C_{1} + C_{2} ln x) x^{0} = C_{1} + C_{2} ln x

— a constant plus

ln x

; this happens for

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

where

c = 0

and

b = a

What happens to the solution structure if

β \to 0

in the complex case

m = α \pm i β

The roots merge into a real repeated root

m = α

; correspondingly

cos (β ln x) \to 1

and

\frac{s i n ( β l n x )}{β} \to ln x

, smoothly recovering the repeated-root form

(C_{1} + C_{2} ln x) x^{α}

x = 0

ever a legal point in the domain of a Cauchy-Euler solution?

Not for solutions like

x^{- 1}

ln x

, or non-integer powers — they are singular at

0

; only when both roots are non-negative integers can a solution stay finite there, but the general solution still excludes

x = 0

because independence typically requires the singular partner.

For

x < 0

, why is

x^{m}

with irrational

m

problematic, and what is the remedy?

A negative base raised to an irrational power is not real-valued, so

x^{m}

is ill-defined; the remedy is

∣ x ∣^{m}

, which solves the same ODE on

x < 0

and stays real.

If a first-order Cauchy-Euler equation

x y^{'} + cy = 0

is given, does the "

ln x

" ever appear?

No repeated-root

ln x

can appear in first order — the indicial is linear (

m + c / a

-style) with a single root;

ln x

arises only when a repeated root demands a second independent solution, which needs order

\geq 2

Can the substitution

t = ln x

be used verbatim on the interval

x < 0

No —

ln x

is undefined there; you must use

t = ln ∣ x ∣

, after which the constant-coefficient machinery works identically on the negative branch.

Does replacing

x^{2} y^{''} + b x y^{'} + cy = 0

by dividing through by

x^{2}

change where the equation is singular?

No — dividing gives

y^{''} + \frac{b}{x} y^{'} + \frac{c}{x ^{2}} y = 0

; the coefficients still blow up at

x = 0

, confirming

x = 0

is a genuine (regular) singular point, not an artefact of how we wrote it.

Recall One-line self-test

The three traps in one breath: middle coefficient is == $(b - a)$ not $b$ , repeated-root partner is $ln x$ not $x$ , and oscillation is in $ln x$ not $x$ ==.