Cauchy–Euler Equations

Almost every solvable second-order ODE we meet has constant coefficients — the same a, b, c multiplying y'', y', y at every point. Let the coefficients vary with x and the door usually slams shut: no elementary closed form. But there is one gorgeous exception — an equation whose variable coefficients are arranged so precisely that it solves as cleanly as any constant-coefficient one, if you spot the pattern. It is the Cauchy–Euler equation (also called the equidimensional or Euler equation):

a\,x^2 y'' + b\,x\,y' + c\,y = 0, \qquad a \ne 0.

Look at the staircase of powers: x^2 rides on the second derivative, x^1 on the first, x^0 = 1 on y itself. Each time you differentiate, you drop a power of x; each coefficient supplies exactly enough x to put it back. That balance — the reason it is called equidimensional — is what makes the whole thing collapse into algebra. Where the constant-coefficient equation rewarded the guess y = e^{rx}, this one rewards a power.

The guess, and the indicial equation

Step 1 — guess a power. What function keeps its shape when you multiply a derivative by the right power of x? A power of x itself. Try

y = x^m, \qquad y' = m\,x^{m-1}, \qquad y'' = m(m-1)\,x^{m-2}.

Step 2 — substitute. Watch each term land back on x^m:

a\,x^2\!\cdot\! m(m-1)x^{m-2} + b\,x\!\cdot\! m\,x^{m-1} + c\,x^m = \big[a\,m(m-1) + b\,m + c\big]x^m = 0.

Step 3 — divide out x^m. For x > 0 it is never zero, so the differential equation becomes a plain quadratic in m — the indicial equation (or auxiliary equation):

\boxed{\,a\,m(m-1) + b\,m + c = 0.\,}

Calculus has dissolved into algebra once more. Expanding gives a m^2 + (b - a)m + c = 0; the m(m-1) form is the one worth memorising, because the -a hiding in the middle term is precisely what people forget. As with the constant-coefficient story, the two roots m_1, m_2 sort into three cases.

Case 1 — distinct real roots

Two different real roots m_1 \ne m_2 give two independent power solutions, and the general solution is their combination:

y = C_1\,x^{m_1} + C_2\,x^{m_2}.

Worked example. Solve x^2 y'' + 2x y' - 6y = 0. Here a=1, b=2, c=-6, so the indicial equation is

m(m-1) + 2m - 6 = m^2 + m - 6 = (m + 3)(m - 2) = 0,

giving m = 2 and m = -3. Hence

y = C_1\,x^{2} + C_2\,x^{-3}.

One solution grows like a parabola, the other blows up near x = 0 and fades as x grows — two completely different shapes, both exact.

A second example. For 2x^2 y'' + 3x y' - y = 0 we have a=2, b=3, c=-1, so 2m(m-1) + 3m - 1 = 2m^2 + m - 1 = (2m - 1)(m + 1) = 0, giving m = \tfrac12 and m = -1. Thus

y = C_1\,x^{1/2} + C_2\,x^{-1} = C_1\sqrt{x} + \frac{C_2}{x}.

Fractional and negative powers are perfectly welcome — another reason the trial is x^m, not an exponential.

Case 2 — a repeated root

When the indicial equation has a single (double) root m, the power x^m is only one solution — we are short a partner. Exactly as multiplying by x rescued the constant-coefficient repeated root, here the second solution is x^m \ln x: multiply by \ln x. Hence

y = \big(C_1 + C_2 \ln x\big)\,x^{m}.

Worked example. Solve x^2 y'' - 3x y' + 4y = 0. With a=1, b=-3, c=4 the indicial equation is

m(m-1) - 3m + 4 = m^2 - 4m + 4 = (m - 2)^2 = 0,

a repeated root m = 2, so

y = \big(C_1 + C_2 \ln x\big)\,x^{2}.

A second example. For x^2 y'' + 5x y' + 4y = 0 we get m(m-1) + 5m + 4 = m^2 + 4m + 4 = (m + 2)^2 = 0, a repeated root m = -2. Forgetting the \ln x would leave only y = C_1 x^{-2} — a one-parameter family, one constant short of the two a second-order equation needs. Restoring it:

y = \big(C_1 + C_2 \ln x\big)\,x^{-2}.

Case 3 — complex roots

A negative discriminant gives a conjugate pair m = \alpha \pm i\beta. The power x^{\alpha + i\beta} is genuine but complex; the identity x^{i\beta} = e^{i\beta \ln x} = \cos(\beta \ln x) + i\sin(\beta \ln x) trades it for two real solutions. Hence

y = x^{\alpha}\big(C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)\big).

The real part \alpha is a power-law envelope; the imaginary part \beta sets an oscillation — but one that speeds up or slows down because it rides on \ln x, not x.

Worked example. For x^2 y'' + x y' + y = 0 we have a=b=c=1, so m(m-1) + m + 1 = m^2 + 1 = 0, giving m = \pm i — so \alpha = 0, \beta = 1 — and

y = C_1 \cos(\ln x) + C_2 \sin(\ln x).

Substitute x = e^{t} (so t = \ln x). By the chain rule the Cauchy–Euler equation in x turns into a constant-coefficient equation in t: for instance a x^2 y'' + b x y' + c y = 0 becomes a\,\ddot y + (b - a)\dot y + c\,y = 0 (dots are d/dt). Its characteristic equation is a r^2 + (b-a)r + c = 0 — the very same indicial equation, with r = m! And its solution y = e^{mt} is just x^m back in the original variable, while the repeated-root partner t\,e^{mt} becomes x^m \ln x and the complex e^{\alpha t}\cos\beta t becomes x^{\alpha}\cos(\beta \ln x). So Cauchy–Euler is the constant-coefficient equation, wearing a change of clothes — which is exactly why the trial power, the \ln x, and the \ln x inside the cosine all appear where they do.

The trial is y = x^m, a power — not e^{mx}. Cauchy–Euler has variable coefficients (x^2, x, 1), so the constant-coefficient guess is wrong here. Reaching for e^{mx} out of habit is the classic slip.
The indicial equation is a\,m(m-1) + b\,m + c = 0, not a m^2 + b m + c = 0. The m(m-1) comes from x^2 y'' = m(m-1)x^m. Expanded it is a m^2 + (b - a)m + c = 0 — note the coefficient of the middle term is b - a, not b. Only when a = 1 do the two forms coincide.
Solutions are usually taken for x > 0. We divided by x^m and wrote \ln x, both of which need x > 0. For x < 0 use |x| (e.g. |x|^m, \ln|x|); the point x = 0 is singular and is excluded.

For a x^2 y'' + b x y' + c y = 0 (x > 0), substitute y = x^m to obtain the indicial equation a\,m(m-1) + b\,m + c = 0. Its roots give the general solution:

Distinct real roots m_1 \ne m_2: \;y = C_1 x^{m_1} + C_2 x^{m_2}.
Repeated root m: \;y = (C_1 + C_2 \ln x)\,x^{m}.
Complex roots m = \alpha \pm i\beta: \;y = x^{\alpha}\big(C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)\big).

The substitution x = e^{t} turns the equation into the constant-coefficient equation a\,\ddot y + (b-a)\dot y + c\,y = 0, whose characteristic equation is the same indicial equation.

See the two basis solutions

For distinct real roots the solution is a blend of two powers, x^{m_1} and x^{m_2}. Slide the two exponents and watch their shapes over x > 0: a positive exponent grows, a negative one blows up near 0 and decays, and m = 0 is a flat line. Set m_1 = 2, m_2 = -3 to see the first worked example's pair; the general solution C_1 x^{m_1} + C_2 x^{m_2} is any weighted mix of these two curves.