Cauchy-Euler (Equidimensional) equation: Worked examples

Every example below tells you WHICH cell of the matrix it fills. Guess the answer first (the "Forecast" line), then check yourself against the worked steps.

The scenario matrix

Every Cauchy-Euler problem is pinned down by (a) the nature of the roots and (b) the "twist" layered on top. Here is the full grid:

Cell	Root type	Twist / degeneracy	Example
A	Distinct real	plain homogeneous	Ex 1
B	Distinct real	initial-value problem (solve for $C_{1}, C_{2}$ )	Ex 2
C	Repeated	$ln x$ appears	Ex 3
D	Complex	oscillation in $ln x$	Ex 4
E	any	domain twist: $x < 0$ , use $	x
F	any	singular point $x = 0$ , limiting behaviour	Ex 6
G	Distinct real	non-homogeneous RHS (variation of parameters)	Ex 7
H	Distinct real	3rd-order (cubic indicial)	Ex 8
I	Complex	real-world word problem (radial physics)	Ex 9

Prerequisites we lean on: Constant-Coefficient Linear ODEs, the Characteristic / Auxiliary Equation, Euler's Formula, Reduction of Order, and Variation of Parameters. The parent is Cauchy-Euler (Equidimensional) equation.

Recall The master recipe (memorise before starting)

For $a x^{2} y^{''} + b x y^{'} + cy = 0$ , guess $y = x^{m}$ and get the indicial equation $am (m - 1) + bm + c = 0$ . Then:

distinct real $m_{1}, m_{2}$ : $y = C_{1} x^{m_{1}} + C_{2} x^{m_{2}}$
repeated $m$ : $y = (C_{1} + C_{2} ln ∣ x ∣) x^{m}$
complex $α \pm i β$ : $y = ∣ x ∣^{α} [C_{1} cos (β ln ∣ x ∣) + C_{2} sin (β ln ∣ x ∣)]$

Cell A — distinct real roots (plain)

Worked example Ex 1 ·

$x^{2} y^{''} + 2 x y^{'} - 6 y = 0$ Forecast: Two roots, two powers. One root will be positive (a growing curve), one negative (a $1/ x^{k}$ curve). Guess before reading which powers.

Step 1. Guess $y = x^{m}$ , so $y^{'} = m x^{m - 1}$ , $y^{''} = m (m - 1) x^{m - 2}$ . Why this step? In an equidimensional equation, $x^{k} \frac{d ^{k}}{d x ^{k}} x^{m}$ is always a constant times $x^{m}$ , so a pure power is the only shape that could survive.

Step 2. Substitute: $x^{2} m (m - 1) x^{m - 2} + 2 x m x^{m - 1} - 6 x^{m} = 0$ , and every $x$ -power collapses to $x^{m}$ : $[m (m - 1) + 2 m - 6] x^{m} = 0.$ Why this step? $x^{2} \cdot x^{m - 2} = x^{m}$ and $x \cdot x^{m - 1} = x^{m}$ ; the whole equation becomes (number) $\times x^{m}$ .

Step 3. For $x \neq = 0$ the bracket must vanish: $m^{2} + m - 6 = 0 \Rightarrow (m + 3) (m - 2) = 0 \Rightarrow m = 2, - 3$ . Why this step? We collected $m (m - 1) + 2 m = m^{2} + m$ ; factoring gives the two exponents.

Step 4. Distinct roots → independent powers → general solution $y = C_{1} x^{2} + C_{2} x^{- 3} .$ Why this step? Two powers with different exponents are linearly independent, so their combination is the full family.

Verify: Plug $y = x^{2}$ : $x^{2} (2) + 2 x (2 x) - 6 x^{2} = 2 x^{2} + 4 x^{2} - 6 x^{2} = 0$ . ✓ Plug $y = x^{- 3}$ : $x^{2} (12 x^{- 5}) + 2 x (- 3 x^{- 4}) - 6 x^{- 3} = 12 x^{- 3} - 6 x^{- 3} - 6 x^{- 3} = 0$ . ✓

Cell B — distinct real + initial-value problem

Worked example Ex 2 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = 0$ , with $y (1) = 3, y^{'} (1) = 0$ Forecast: From the parent note the roots are $m = 4, - 1$ . The initial data must pick specific $C_{1}, C_{2}$ . Since $y^{'} (1) = 0$ (flat start), expect the two powers to balance their slopes at $x = 1$ .

Step 1. General solution (parent Ex): $y = C_{1} x^{4} + C_{2} x^{- 1}$ . Why this step? Indicial $m^{2} - 3 m - 4 = (m - 4) (m + 1) = 0$ .

Step 2. Apply $y (1) = 3$ : $C_{1} (1) + C_{2} (1) = 3 \Rightarrow C_{1} + C_{2} = 3$ . Why this step? At $x = 1$ both powers equal $1$ , giving a simple sum.

Step 3. $y^{'} = 4 C_{1} x^{3} - C_{2} x^{- 2}$ ; apply $y^{'} (1) = 0$ : $4 C_{1} - C_{2} = 0 \Rightarrow C_{2} = 4 C_{1}$ . Why this step? We need a second equation; differentiate then evaluate.

Step 4. Solve: $C_{1} + 4 C_{1} = 3 \Rightarrow C_{1} = \frac{3}{5}, C_{2} = \frac{12}{5}$ . Why this step? Two linear equations, two unknowns. $y = \frac{3}{5} x^{4} + \frac{12}{5} x^{- 1} .$

Verify: $y (1) = \frac{3}{5} + \frac{12}{5} = \frac{15}{5} = 3$ ✓. $y^{'} (1) = 4 \cdot \frac{3}{5} - \frac{12}{5} = \frac{12}{5} - \frac{12}{5} = 0$ ✓.

Cell C — repeated root ( $ln x$ appears)

Worked example Ex 3 ·

$x^{2} y^{''} - 5 x y^{'} + 9 y = 0$ Forecast: If the discriminant of the indicial quadratic is zero, we get ONE power and must sprinkle in $ln x$ for the partner. Guess: does the second curve grow faster or slower than $x^{m}$ ? (Look at the orange curve in the figure — $ln x$ grows, so it grows slightly faster.)

Step 1. Indicial: $m (m - 1) - 5 m + 9 = m^{2} - 6 m + 9 = (m - 3)^{2} = 0 \Rightarrow m = 3$ (double). Why this step? $b - a = - 5 - 1 = - 6$ ; the perfect square signals a repeated root.

Step 2. One solution is $y_{1} = x^{3}$ . The second comes from the $x = e^{t}$ picture: a repeated root there gives $e^{3 t}$ AND $t e^{3 t}$ ; translating $t = ln x$ gives $(ln x) x^{3}$ . Why this step? This is Reduction of Order in disguise — the $t$ -world hands us the missing partner automatically.

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

Verify: Let $y = x^{3} ln x$ . Then $y^{'} = 3 x^{2} ln x + x^{2}$ , $y^{''} = 6 x ln x + 5 x$ . Substitute: $x^{2} (6 x ln x + 5 x) - 5 x (3 x^{2} ln x + x^{2}) + 9 x^{3} ln x = 6 x^{3} ln x + 5 x^{3} - 15 x^{3} ln x - 5 x^{3} + 9 x^{3} ln x = 0$ ✓ (both the $ln$ terms and the plain terms cancel).

Cell D — complex roots (oscillation in $ln x$ )

Worked example Ex 4 ·

$x^{2} y^{''} + 3 x y^{'} + 5 y = 0$ Forecast: A negative discriminant means the roots spin into the complex plane. The solution won't be a plain power — it will oscillate, but in $ln x$ , so the wiggles stretch out as $x$ grows (see the compressing-then-stretching wave in the figure). The real part $α$ sets whether the envelope grows or decays.

Step 1. Indicial: $m (m - 1) + 3 m + 5 = m^{2} + 2 m + 5 = 0$ . Why this step? $b - a = 3 - 1 = 2$ , $c = 5$ .

Step 2. Quadratic formula: $m = \frac{- 2 \pm 4 - 20}{2} = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i$ . So $α = - 1, β = 2$ . Why this step? $- 16 = 4 i$ ; we split into real part $α$ and imaginary part $β$ .

Step 3. Use $x^{α \pm i β} = x^{α} e^{\pm i β l n x}$ and Euler's Formula $e^{i θ} = cos θ + i sin θ$ with $θ = β ln x$ : $y = x^{- 1} [C_{1} cos (2 ln x) + C_{2} sin (2 ln x)] .$ Why this step? Euler converts the complex power into real oscillations; $x^{- 1}$ is a decaying envelope so the wiggles shrink as $x \to \infty$ .

Verify: Take $y = x^{- 1} cos (2 ln x)$ . Numerically at $x = e$ (so $ln x = 1$ ): direct substitution of $y, y^{'}, y^{''}$ into the ODE gives $0$ — confirmed in the machine check.

Cell E — domain twist: $x < 0$ , use $∣ x ∣$

Worked example Ex 5 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = 0$ on $x < 0$ Forecast: Same equation as parent Ex 1 (roots $m = 4, - 1$ ) but now $x$ is negative. Raising a negative number to $- 1$ is fine, but the general theory wants $∣ x ∣$ . Guess: does the answer look identical or does a sign creep in?

Step 1. Roots are still $m = 4, - 1$ (the indicial equation never saw $x$ 's sign). Why this step? The indicial equation is pure algebra in $m$ .

Step 2. For $x < 0$ replace each power $x^{m}$ by $∣ x ∣^{m}$ : $y = C_{1} ∣ x ∣^{4} + C_{2} ∣ x ∣^{- 1} .$ Why this step? When $m$ is non-integer, $x^{m}$ isn't real for $x < 0$ ; using $∣ x ∣^{m}$ keeps solutions real on the whole negative ray. (Here $∣ x ∣^{4} = x^{4}$ but $∣ x ∣^{- 1} = - x^{- 1}$ for $x < 0$ , so the constant absorbs the sign.)

Step 3. Because $ln x$ is undefined for $x < 0$ , any repeated/complex case would use $ln ∣ x ∣$ instead. Here that doesn't arise, but keep the rule: always $∣ x ∣$ off the positive ray.

Verify: Take $x = - 2$ , $y = ∣ x ∣^{- 1} = \frac{1}{2}$ . Then $y = - x^{- 1}$ , $y^{'} = x^{- 2}$ , $y^{''} = - 2 x^{- 3}$ . Substitute at general $x < 0$ : $x^{2} (- 2 x^{- 3}) - 2 x (x^{- 2}) - 4 (- x^{- 1}) = - 2 x^{- 1} - 2 x^{- 1} + 4 x^{- 1} = 0$ ✓.

Cell F — the singular point $x = 0$ (limiting behaviour)

Worked example Ex 6 · Behaviour of

$x^{2} y^{''} + 2 x y^{'} - 6 y = 0$ as $x \to 0^{+}$ Forecast: From Ex 1 the solution is $C_{1} x^{2} + C_{2} x^{- 3}$ . At $x = 0$ the coefficient of $y^{''}$ (namely $x^{2}$ ) vanishes — this is a regular singular point (Frobenius Method & Regular Singular Points). Guess which piece explodes and which stays bounded.

Step 1. Write $y = C_{1} x^{2} + C_{2} x^{- 3}$ and take $x \to 0^{+}$ . Why this step? Limits reveal which basis solution is "physical" near the singular point.

Step 2. $x^{2} \to 0$ (bounded, in fact vanishing) but $x^{- 3} \to + \infty$ . Why this step? A positive exponent decays into the origin; a negative exponent blows up. The larger the $∣ m ∣$ , the more violent the blow-up.

Step 3. For a solution to stay bounded at the origin we must set $C_{2} = 0$ , leaving $y = C_{1} x^{2}$ . Why this step? Physical problems on a disk (temperature at the centre, no source) reject infinities, selecting the regular branch.

Verify: At $x = 0.1$ : $x^{2} = 0.01$ (tiny), $x^{- 3} = 1000$ (huge). Ratio $1 0^{5}$ confirms $x^{- 3}$ dominates near $0$ . As $x \to 0^{+}$ , $x^{- 3} \to \infty$ ✓.

Cell G — non-homogeneous (variation of parameters)

Worked example Ex 7 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = x^{3}$ on $x > 0$ Forecast: Homogeneous part is Ex 1 ( $C_{1} x^{4} + C_{2} x^{- 3}$ ). The forcing $x^{3}$ is itself a power, so expect a particular solution $\propto x^{3}$ — unless $x^{3}$ collides with a homogeneous power (it doesn't; $3 \neq = 4, - 3$ ).

Step 1. $y_{h} = C_{1} x^{4} + C_{2} x^{- 3}$ ; try $y_{p} = A x^{3}$ . Why this step? The RHS is a power, and no homogeneous root equals $3$ , so an undetermined-coefficient power works. (For a colliding power you'd multiply by $ln x$ ; see Variation of Parameters for the general method.)

Step 2. $y_{p}^{'} = 3 A x^{2}$ , $y_{p}^{''} = 6 A x$ . Substitute: $x^{2} (6 A x) - 2 x (3 A x^{2}) - 4 A x^{3} = 6 A x^{3} - 6 A x^{3} - 4 A x^{3} = - 4 A x^{3} .$ Why this step? Every term is a multiple of $x^{3}$ ; collect them.

Step 3. Match to RHS $x^{3}$ : $- 4 A = 1 \Rightarrow A = - \frac{1}{4}$ . So $y_{p} = - \frac{1}{4} x^{3}$ . Why this step? The coefficient of $x^{3}$ on the left must equal $1$ .

Step 4. $y = C_{1} x^{4} + C_{2} x^{- 3} - \frac{1}{4} x^{3} .$

Verify: Substitute $y_{p} = - \frac{1}{4} x^{3}$ : LHS $= - 4 (- \frac{1}{4}) x^{3} = x^{3} =$ RHS ✓.

Cell H — third order (cubic indicial)

Worked example Ex 8 ·

$x^{3} y^{'''} + 2 x^{2} y^{''} - 4 x y^{'} + 4 y = 0$ Forecast: The general $n$ -th order Cauchy-Euler still yields a polynomial in $m$ — this time a cubic. Expect three exponents. Guess whether any are equal.

Step 1. For $y = x^{m}$ : $y^{'} = m x^{m - 1}$ , $y^{''} = m (m - 1) x^{m - 2}$ , $y^{'''} = m (m - 1) (m - 2) x^{m - 3}$ . Why this step? Same equidimensional trick; $x^{3} \cdot x^{m - 3} = x^{m}$ , etc.

Step 2. Indicial: $m (m - 1) (m - 2) + 2 m (m - 1) - 4 m + 4 = 0.$ Expand: $m (m - 1) (m - 2) = m^{3} - 3 m^{2} + 2 m$ and $2 m (m - 1) = 2 m^{2} - 2 m$ , so $m^{3} - 3 m^{2} + 2 m + 2 m^{2} - 2 m - 4 m + 4 = m^{3} - m^{2} - 4 m + 4 = 0.$ Why this step? Collect all powers of $m$ into one cubic.

Step 3. Factor: test $m = 1$ : $1 - 1 - 4 + 4 = 0$ ✓. Divide out $(m - 1)$ : $m^{3} - m^{2} - 4 m + 4 = (m - 1) (m^{2} - 4) = (m - 1) (m - 2) (m + 2)$ . Roots $m = 1, 2, - 2$ . Why this step? Rational-root guess then factor the remaining quadratic.

Step 4. Three distinct real roots → three independent powers: $y = C_{1} x + C_{2} x^{2} + C_{3} x^{- 2} .$

Verify: Plug $y = x^{2}$ : $x^{3} (0) + 2 x^{2} (2) - 4 x (2 x) + 4 x^{2} = 0 + 4 x^{2} - 8 x^{2} + 4 x^{2} = 0$ ✓. Plug $y = x^{- 2}$ : $x^{3} (- 24 x^{- 5}) + 2 x^{2} (6 x^{- 4}) - 4 x (- 2 x^{- 3}) + 4 x^{- 2} = - 24 x^{- 2} + 12 x^{- 2} + 8 x^{- 2} + 4 x^{- 2} = 0$ ✓.

Cell I — real-world word problem (radial physics)

Worked example Ex 9 · Steady temperature in a hollow disk

The steady-state temperature $u (r)$ in a thin annular plate (with an angular variation) can satisfy $r^{2} u^{''} + r u^{'} - 4 u = 0, 1 \leq r \leq e, u (1) = 0, u (e) = 10.$ Forecast: Radial equations from Laplace's operator are always Cauchy-Euler (the $r^{2}$ , $r$ , constant pattern is baked into the geometry). Here $c = - 4 < 0$ gives distinct real roots $\pm 2$ : a growing and a decaying radial mode.

Step 1. Indicial: $m (m - 1) + m - 4 = m^{2} - 4 = 0 \Rightarrow m = \pm 2$ . Why this step? $b - a = 1 - 1 = 0$ , so no linear $m$ term; just $m^{2} - 4$ .

Step 2. General radial solution: $u = C_{1} r^{2} + C_{2} r^{- 2}$ . Why this step? Distinct real roots → two power modes.

Step 3. Apply $u (1) = 0$ : $C_{1} + C_{2} = 0 \Rightarrow C_{2} = - C_{1}$ . Apply $u (e) = 10$ : $C_{1} e^{2} + C_{2} e^{- 2} = 10 \Rightarrow C_{1} (e^{2} - e^{- 2}) = 10$ . Why this step? Boundary conditions pin the constants.

Step 4. $C_{1} = \frac{10}{e ^{2} - e ^{- 2}}$ , $C_{2} = - C_{1}$ . Numerically $e^{2} - e^{- 2} \approx 7.2537$ , so $C_{1} \approx 1.3786$ , $C_{2} \approx - 1.3786$ . $u (r) = 1.3786 (r^{2} - r^{- 2}) (approx) .$

Verify: $u (1) = 1.3786 (1 - 1) = 0$ ✓. $u (e) = 1.3786 (e^{2} - e^{- 2}) = 1.3786 \times 7.2537 \approx 10$ ✓. Units: $u$ in °C, $r$ dimensionless (scaled radius), each term $r^{\pm 2}$ dimensionless × °C — consistent.