Euler-Cauchy equations

Most linear ODEs with variable coefficients have no closed-form solution at all — you fall back on series methods or numerical integration. The Euler-Cauchy equation is the striking exception the MA10001 syllabus singles out: a variable-coefficient equation, $x^2y'' + bxy' + cy = f(x)$ , that is nonetheless solvable exactly, by a single clean substitution. The trick is $x = e^t$ . It looks like an arbitrary change of variable, but it is doing something precise: it converts multiplication by $x$ into differentiation with respect to $t$ , so a power-of- $x$ coefficient pattern collapses into a constant-coefficient ODE — the same auxiliary-equation machinery you already know from the constant-coefficient unit. This article expands the Pass-1 treatment with a formal derivation of the operator identity $x^2y''=D(D-1)y$ , a fuller worked example, competitive-exam shortcuts, and a compact exam card. Every number and result below matches the Pass-1 article exactly; nothing new is claimed.

The core insight

The substitution $x = e^t$ turns the equidimensional equation $x^2y'' + bxy' + cy = f(x)$ into a constant-coefficient ODE in $t$ , because $x\,\dfrac{dy}{dx} = Dy$ and $x^2\dfrac{d^2y}{dx^2} = D(D-1)y$ with $D \equiv d/dt$ .

The core idea

A second-order Euler-Cauchy (equidimensional) equation looks like

x^2\frac{d^2y}{dx^2} + bx\frac{dy}{dx} + cy = f(x), \qquad x > 0,

where: $b, c$ are constants and $f(x)$ is the (possibly zero) forcing term. The defining feature is that the power of $x$ multiplying each derivative exactly matches its order: $x^2$ pairs with $y''$ , $x^1$ with $y'$ , $x^0$ with $y$ . That matching is what makes the equation "equidimensional" — scale $x$ by any constant and the equation keeps its shape.

Why $x = e^t$ works. Set $t = \ln x$ and write $D \equiv d/dt$ . By the chain rule, $\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{1}{x}Dy$ , so

x\frac{dy}{dx} = Dy.

Differentiating again and simplifying (done in full below) gives the second-order identity

x^2\frac{d^2y}{dx^2} = D(D-1)y = D^2y - Dy.

where: $D(D-1)$ means "apply $D$ , then apply $(D-1)$ " — it is a product of operators, not of numbers, but algebraically it expands the same way. Substituting both identities into the Euler-Cauchy equation removes every explicit $x$ , leaving a constant-coefficient linear ODE in $t$ :

D^2y + (b-1)Dy + cy = f(e^t).

That is solved with the standard toolkit: form the auxiliary equation in $m$ (treat $D \to m$ ), find the complementary function, add a particular integral if $f \neq 0$ , then substitute back $t = \ln x$ . Because $e^{mt} = e^{m\ln x} = x^m$ , the root types translate directly: real distinct roots $m_1, m_2$ give $x^{m_1}, x^{m_2}$ ; a repeated root $m$ gives $x^m$ and $x^m\ln x$ ; complex roots $\alpha \pm i\beta$ give $x^\alpha\cos(\beta\ln x)$ and $x^\alpha\sin(\beta\ln x)$ . Third- and higher-order Euler-Cauchy equations extend the same pattern, with $x^3y''' \to D(D-1)(D-2)y$ and so on. There is a second, equivalent route that skips $t$ altogether: try $y = x^m$ directly in the homogeneous equation $x^2y'' + bxy' + cy = 0$ . Substituting gives $x^m[m(m-1) + bm + c] = 0$ , so $m(m-1) + bm + c = 0$ — identical to the auxiliary equation obtained via $D(D-1) + bD + c$ . Either derivation is acceptable on this course, but the $x = e^t$ substitution is preferred here because it also carries the nonhomogeneous case cleanly into the constant-coefficient particular-integral machinery.

Figure

Diagram of the Euler-Cauchy substitution mapping x squared y double prime plus b x y prime plus c y equals f of x through x equals e to the t into a constant coefficient equation D squared y plus b minus 1 D y plus c y equals f of e to the t, then back through t equals ln x to the final solution in x

Derivation of the operator identity

Derivation ADeriving x squared y double prime equals D of D minus one y

The goal is to prove, starting only from the substitution $t = \ln x$ and the ordinary chain rule, that $x\dfrac{dy}{dx} = Dy$ and $x^2\dfrac{d^2y}{dx^2} = D(D-1)y = D^2y - Dy$ , so that every Euler-Cauchy equation converts exactly into a constant-coefficient equation in $t$ with no approximation anywhere in the argument.

Step 1. Set $t = \ln x$ , so $\dfrac{dt}{dx} = \dfrac{1}{x}$ . By the chain rule, $\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{1}{x}\dfrac{dy}{dt} = \dfrac{1}{x}Dy$ . Multiplying both sides by $x$ gives the first-derivative identity immediately: $x\dfrac{dy}{dx} = Dy$ .

Step 2. Differentiate $\dfrac{dy}{dx} = \dfrac{1}{x}Dy$ once more with respect to $x$ , treating $Dy$ (a function of $t$ , hence of $x$ ) with the product rule. $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{1}{x}Dy\right) = -\dfrac{1}{x^2}Dy + \dfrac{1}{x}\dfrac{d(Dy)}{dx}$ .

Step 3. Expand $\dfrac{d(Dy)}{dx}$ by the chain rule again, exactly as in Step 1 but applied to the function $Dy$ instead of $y$ : $\dfrac{d(Dy)}{dx} = \dfrac{1}{x}\dfrac{d(Dy)}{dt} = \dfrac{1}{x}D(Dy) = \dfrac{1}{x}D^2y$ .

Step 4. Substitute back into Step 2: $\dfrac{d^2y}{dx^2} = -\dfrac{1}{x^2}Dy + \dfrac{1}{x}\cdot\dfrac{1}{x}D^2y = \dfrac{1}{x^2}\big(D^2y - Dy\big)$ . Multiplying both sides by $x^2$ clears the fraction and gives $x^2\dfrac{d^2y}{dx^2} = D^2y - Dy = D(D-1)y$ , exactly the identity used to convert the Euler-Cauchy equation. Every $x^2y''$ term in the original equation can now be replaced by $D(D-1)y$ , and every $xy'$ term by $Dy$ , turning the whole left side into a polynomial in $D$ with constant coefficients.

Worked example

Solve the homogeneous Euler-Cauchy equation $x^2y'' - 2xy' - 4y = 0$ , then verify the solution directly.

Step 1. Substitute $x = e^t$ . Using $x^2y'' = D(D-1)y$ and $xy' = Dy$ , the equation becomes $D(D-1)y - 2Dy - 4y = 0$ , which expands to $D^2y - Dy - 2Dy - 4y = 0$ , i.e. $D^2y - 3Dy - 4y = 0$ — a constant-coefficient equation in $t$ .

Step 2. Form the auxiliary equation in $m$ : $m^2 - 3m - 4 = 0$ . Factoring, $(m-4)(m+1) = 0$ , so $m = 4$ or $m = -1$ — two real, distinct roots.

Step 3. Write the general solution in $t$ using the two real roots: $y = C_1e^{4t} + C_2e^{-t}$ , where $C_1, C_2$ are arbitrary constants fixed by initial or boundary conditions.

Step 4. Substitute back $t = \ln x$ , using $e^{mt} = x^m$ : $e^{4t} = x^4$ and $e^{-t} = x^{-1}$ , so $y = C_1x^4 + C_2x^{-1}$ .

Step 5 — verify by direct substitution. For $y = x^4$ : $y' = 4x^3$ , $y'' = 12x^2$ . Then $x^2(12x^2) - 2x(4x^3) - 4x^4 = 12x^4 - 8x^4 - 4x^4 = 0$ , confirming $y = x^4$ solves the equation; a matching check on $y = x^{-1}$ confirms the second term. The general solution is $y = C_1x^4 + C_2x^{-1}$ .

Competitive stretch

For the nonhomogeneous case $x^2y'' + xy' - y = x^2$ ( $x>0$ ), the same machine runs with an extra particular-integral step: $x=e^t$ turns the left side into $D^2y - y$ and the right side $f(x)=x^2$ into $f(e^t)=(e^t)^2=e^{2t}$ , giving $D^2y-y=e^{2t}$ . The auxiliary equation $m^2-1=0$ gives $m=\pm1$ , so $y_c=C_1e^t+C_2e^{-t}$ . Since $e^{2t}$ does not clash with the roots $\pm1$ , try $y_p=Ae^{2t}$ : substituting, $4Ae^{2t}-Ae^{2t}=3Ae^{2t}=e^{2t}$ , so $A=\tfrac13$ . The full solution in $t$ is $y=C_1e^t+C_2e^{-t}+\tfrac13e^{2t}$ , and substituting back $t=\ln x$ (so $e^t=x$ , $e^{-t}=1/x$ , $e^{2t}=x^2$ ) gives $y = C_1x + \dfrac{C_2}{x} + \dfrac{x^2}{3}$ . A JEE-style question will often just ask for the particular integral, so recognizing $f(e^t)=e^{2t}$ and checking it against the roots of the auxiliary equation is the fast path — no need to re-derive the complementary function from scratch if it is already given.

Common traps

Using $x^2y'' \to D^2y$ instead of the correct $D(D-1)y = D^2y - Dy$ . The missing $-Dy$ term, traced in the derivation above, is the single most common Euler-Cauchy slip.
Forgetting to convert $f(x)$ into a function of $t$ . $f(x)=x^3$ becomes $f(e^t)=e^{3t}$ ; leaving it as $x^3$ on the right side of a $D$ -operator equation is inconsistent and makes undetermined coefficients unusable.
Forgetting to substitute back $t = \ln x$ at the end, leaving the final answer as a function of $t$ instead of $x$ .
Mishandling repeated or complex roots after transforming back. A repeated root $m$ gives $x^m$ and $x^m\ln x$ — not $x^m$ and $x\cdot x^m$ ; complex roots $\alpha\pm i\beta$ give $x^\alpha\cos(\beta\ln x)$ and $x^\alpha\sin(\beta\ln x)$ , with $\ln x$ inside the trig functions.
Applying the method to a non-equidimensional equation. It only works when the power of $x$ exactly matches the derivative order at every term; $x^3y''+xy'+y=0$ is not Euler-Cauchy and needs a different (out-of-syllabus) approach.

Ask the tutor: unsure how to identify whether a given ODE is genuinely Euler-Cauchy, or want a step-by-step check of your $D(D-1)$ substitution on a new problem — ask the AI tutor to walk through it with you.

Why does x = e^t work instead of some other substitution?

Because it exactly cancels the pattern of the equation. Each Euler-Cauchy term $x^ky^{(k)}$ has its power of $x$ matched to its derivative order, and under $x=e^t$ every derivative with respect to $x$ picks up a compensating $1/x$ factor that combines with the $x^k$ coefficient to leave a pure operator in $D=d/dt$ with no $x$ or $t$ left explicitly. No other simple substitution produces this exact cancellation for all orders at once, which is why $x=e^t$ is the standard choice rather than, say, $x=t^2$ or $x=\sin t$ .

Do I have to go through t, or can I just guess y = x^m directly?

Both routes are valid and give the same auxiliary equation. Trying $y=x^m$ in the homogeneous equation directly gives $x^m[m(m-1)+bm+c]=0$ , so $m(m-1)+bm+c=0$ — identical to what $D(D-1)+bD+c$ produces via $x=e^t$ . The direct guess is faster for homogeneous equations, but the $x=e^t$ route is preferred once the equation is nonhomogeneous, because it lets you reuse the full undetermined-coefficients or variation-of-parameters toolkit you already know for constant-coefficient equations, applied to $f(e^t)$ on the right side.

What if the coefficient powers of x don't line up with the derivative orders?

Then the equation is not Euler-Cauchy and this substitution will not fully clear the $x$ -dependence. For example, $x^3y''+xy'+y=0$ has $x^3$ multiplying $y''$ instead of the required $x^2$ , so $x=e^t$ leaves a residual $x$ term behind. Equations like this need a different method (often a series solution about $x=0$ ), which lies outside this unit's scope — recognizing the mismatched powers before attempting the substitution is the key skill, not memorizing the alternative method.