The Cauchy-Euler Equation

Quick Reference

Item	Statement
General form	$a_n(x-x_0{)}^n{y}^{(n)}+a_{n-1}(x-x_0{)}^{n-1}{y}^{(n-1)}+\cdots +a_{0}y=0$, $x\ne x_0$
Second-order form	ax^{2}y'' + bxy' + cy = 0,
Substitution 1	$x-x_0=e^z$, that is, $z=\ln (x-x_0)$; gives constant coefficients
Substitution 2	$y=(x-x_0{)}^r$; gives an algebraic equation in $r$
Characteristic equation	$ar(r-1)+br+c=0$, that is, $a{r}^2+(b-a)r+c=0$
Distinct real roots	$y=c_1{x}^{r_1}+c_2{x}^{r_2}$
Repeated real root	$y=c_1{x}^{r_1}+c_2{x}^{r_1}\ln x$
Complex roots $\alpha \pm i\beta$	$y=x^{\alpha }[c_1\cos (\beta \ln x)+c_2\sin (\beta \ln x)]$
Root $r_1$ of multiplicity $k$	$x^{r_1},x^{r_1}\ln x,\dots ,x^{r_1}{(\ln x)}^{k-1}$
For	Replace $x$ by $\left|x\right|$ throughout

Definition

This is one of the few linear differential equations with variable coefficients that has a closed-form solution.

For the rest of this section we take $x_0=0$, which loses no generality since the substitution $x\mapsto x-x_0$ recovers the general case.

How to Solve the Cauchy-Euler Equation

Note the parallel with constant coefficients: there, a repeated root produced extra factors of $x$; here it produces extra factors of $\ln x$. The substitution $z=\ln x$ explains why, since $x$ in the $z$-world is $\ln x$ in the $x$-world.

Method 1: Substituting $x=e^z$

Consider the second-order equation

ax^{2}y'' + bxy' + cy = 0, \qquad (x > 0) . \tag{iv}

Introduce a new independent variable $z$ by $x=e^z$, equivalently $z=\ln x$, so that ${\displaystyle \frac{dz}{dx}}={\displaystyle \frac{1}{x}}$ and ${\displaystyle \frac{d^{2}z}{d{x}^2}}=-{\displaystyle \frac{1}{x^2}}$.

By the chain rule,

y' = \frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \frac{1}{x}\frac{dy}{dz}

and, differentiating again with the product and chain rules,

\begin{aligned} y'' = \frac{d^{2}y}{dx^{2}} &= \frac{d}{dx}\left(\frac{dy}{dz}\frac{dz}{dx}\right) \\ &= \left(\frac{d}{dx}\frac{dy}{dz}\right)\frac{dz}{dx} + \frac{dy}{dz}\frac{d^{2}z}{dx^{2}} \\ &= \frac{d^{2}y}{dz^{2}}\left(\frac{dz}{dx}\right)^{2} + \frac{dy}{dz}\frac{d^{2}z}{dx^{2}} \\ &= \frac{1}{x^{2}}\frac{d^{2}y}{dz^{2}} - \frac{1}{x^{2}}\frac{dy}{dz} \end{aligned}

Substituting these expressions into (iv), the factors of $x^2$ and $x$ cancel exactly against the coefficients, which is the whole point of the equidimensional structure:

$$a\frac{d^{2}y}{d{z}^2}+(b-a)\frac{dy}{dz}+cy=0,$$

a linear differential equation with constant coefficients. Its characteristic equation is

$$ar(r-1)+br+c=0\text{or}a{r}^2+(b-a)r+c=0.$$

This may have two distinct real roots $r_1$ and $r_2$, a double root $r_1$, or complex roots $\alpha \pm i\beta$.

The Three Cases

1. Two distinct real roots. Two linearly independent solutions are

$$y_1=e^{r_{1}z}={\left(e^z\right)}^{r_1}=x^{r_1}\text{and}{y}_2=e^{r_{2}z}=x^{r_2}.$$

2. Repeated real root. Two linearly independent solutions are

$$y_1=e^{r_{1}z}=x^{r_1}\text{and}{y}_2=z{y}_1=x^{r_1}\ln x.$$

The extra factor $z$ from the constant-coefficient theory becomes $\ln x$ on translating back.

3. Complex roots $\alpha \pm i\beta$. Two linearly independent solutions are

$$y_1=e^{\alpha z}\cos \beta z=x^{\alpha }\cos (\beta \ln x)\text{and}{y}_2=e^{\alpha z}\sin \beta z=x^{\alpha }\sin (\beta \ln x).$$

Method 2: Substituting $y=x^r$

We may instead assume the solution has the form $y=x^r$ and find $r$ directly. Substituting $y=x^r$, y' = rx^{r-1}, y'' = r(r-1)x^{r-2} into (iv),

\begin{aligned} L\left[x^{r}\right] &= ax^{2}\cdot r(r-1)x^{r-2} + bx\cdot rx^{r-1} + cx^{r} \\ &= \left[ar(r-1) + br + c\right]x^{r} = 0 \end{aligned}

Every term produced $x^r$, again because the powers of $x$ match the orders of differentiation. Since $x^r\ne 0$ for , we conclude that $y=x^r$ is a solution of (iv) whenever $r$ is a root of the characteristic equation

$$p(r)=ar(r-1)+br+c=0,\text{that is,}p(r)=a{r}^2+(b-a)r+c=0.$$

The Three Cases Again

1. Two distinct real roots. If $r_1\ne r_2$ are the roots, then

$$y_1(x)=x^{r_1},y_2(x)=x^{r_2}$$

are two linearly independent solutions of (iv).

2. Repeated real root. If $r_1=r_2$, one solution is $y_1=x^{r_1}$. To construct the second, differentiate both sides of $L\left[x^r\right]=p(r)x^r$ with respect to $r$:

\frac{\partial}{\partial r}L\left[x^{r}\right] = L\left[\frac{\partial}{\partial r}x^{r}\right] = L\left[x^{r}\ln x\right] = p'(r)x^{r} + p(r)x^{r}\ln x .

Recall that ${\displaystyle \frac{d}{du}}{a}^u=a^u\ln a$; in ${\displaystyle \frac{\partial }{\partial r}}{x}^r$ the variable is $r$ and $x$ is held fixed, so the derivative is $x^r\ln x$.

For $r=r_1$ we have p(r_1) = p'(r_1) = 0, since $r_1$ is a double root, and therefore

L\left[x^{r_1}\ln x\right] = p'(r_1)x^{r_1} + p(r_1)x^{r_1}\ln x = 0 .

This means $y_2=x^{r_1}\ln x$ is a solution. So the two linearly independent solutions are

$$y_1(x)=x^{r_1},y_2(x)=x^{r_1}\ln x.$$

3. Complex roots. If the roots of $p(r)=0$ are the conjugate pair $r_1=\alpha +i\beta$ and $r_2=\alpha -i\beta$, the general solution is

$$y=A{x}^{\alpha +i\beta }+B{x}^{\alpha -i\beta },$$

where $A$ and $B$ are constants. Because

\begin{aligned} x^{\alpha+i\beta} &= e^{\ln x^{\alpha+i\beta}} \\ &= e^{(\alpha+i\beta)\ln x} = e^{\alpha\ln x + i\beta\ln x} \qquad \left(\ln A^{B} = B\ln A\right) \\ &= e^{\alpha\ln x}\left[\cos(\beta\ln x) + i\sin(\beta\ln x)\right] \qquad \left(e^{i\theta} = \cos\theta + i\sin\theta\right) \\ &= x^{\alpha}\left[\cos(\beta\ln x) + i\sin(\beta\ln x)\right] \end{aligned}

and similarly

$$x^{\alpha -i\beta }=x^{\alpha }[\cos (\beta \ln x)-i\sin (\beta \ln x)],$$

the general solution of (iv) can be written as

$$y=c_1{x}^{\alpha }\cos (\beta \ln x)+c_2{x}^{\alpha }\sin (\beta \ln x),$$

where $c_1=A+B$ and $c_2=i(A-B)$.

In summary, the two linearly independent solutions are

What If ?

So far we have considered . When , $\ln x$ is undefined. In that case start with the change of independent variable $t=-x$, so that . Then

\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\frac{dt}{dx} = -\frac{dy}{dt} \\ \frac{d^{2}y}{dx^{2}} &= \frac{d^{2}y}{dt^{2}}\left(\frac{dt}{dx}\right)^{2} + \frac{dy}{dt}\frac{d^{2}t}{dx^{2}} = \frac{d^{2}y}{dt^{2}} \end{aligned}

since ${\displaystyle \frac{dt}{dx}}=-1$ and ${\displaystyle \frac{d^{2}t}{d{x}^2}}=0$. Substituting, and noting that $x^2=t^2$ while $x{\displaystyle \frac{dy}{dx}}=(-t)(-{\displaystyle \frac{dy}{dt}})=t{\displaystyle \frac{dy}{dt}}$, the differential equation becomes

which is the same equation in the new variable. Its solution is already known, and replacing $t$ by $-x$ expresses the answer in terms of $x$.

Combining the results for and , the solution of

$$a{x}^2\frac{d^{2}y}{d{x}^2}+bx\frac{dy}{dx}+cy=0$$

on any interval not containing the origin depends on the roots $r_1$, $r_2$ of $p(r)=ar(r-1)+br+c=0$:

In short: put absolute value bars around every $x$. The origin is a singular point of the equation, so solutions are never guaranteed to extend across it.

Worked Examples

Exercises

Frequently Asked Questions

How do I recognize a Cauchy-Euler equation?

Each term pairs $x^k$ with $y^{(k)}$: the power of $x$ equals the order of the derivative. So x^{2}y'', xy', and $y$ all fit, while x^{3}y'' or x^{2}y' do not. The powers must match term by term.

Why does $y=x^r$ work?

Because the equation is invariant under rescaling $x\mapsto \lambda x$, and power functions are the functions that transform simply under rescaling. Concretely, $x^k{\displaystyle \frac{d^k}{d{x}^k}}{x}^r$ is always a constant times $x^r$, so every term produces $x^r$ and the equation collapses to a polynomial condition on $r$.

Why is the characteristic equation $ar(r-1)+br+c$ rather than $a{r}^2+br+c$?

Because ${\displaystyle \frac{d^2}{d{x}^2}}{x}^r=r(r-1)x^{r-2}$, so the second-derivative term contributes $r(r-1)$, not $r^2$. Substituting the trial solution rather than writing down a memorized formula avoids this error entirely.

Why does a repeated root produce $\ln x$ instead of $x$?

The substitution $z=\ln x$ turns the equation into a constant-coefficient one, where a repeated root produces the extra factor $z$. Translating back, $z$ is $\ln x$. Equivalently, differentiating $L\left[x^r\right]=p(r)x^r$ with respect to $r$ produces $x^r\ln x$, since ${\displaystyle \frac{\partial }{\partial r}}{x}^r=x^r\ln x$.

What happens at $x=0$?

It is a singular point: dividing by $x^2$ to reach standard form produces coefficients that blow up there. Solutions such as $x^{-3}$ or $x^2\ln x$ misbehave at the origin, and no solution is guaranteed to extend across it. Always solve on an interval that excludes $x=0$.

Which method should I use, $x=e^z$ or $y=x^r$?

For a homogeneous equation, $y=x^r$ is faster: one substitution and a polynomial to factor. The substitution $x=e^z$ is preferable when the equation is nonhomogeneous, because it converts the problem to a constant-coefficient one where undetermined coefficients becomes available. Alternatively, keep $y=x^r$ for the homogeneous part and finish with variation of parameters.

Does the method extend to higher order?

Yes, unchanged. Substituting $y=x^r$ into $x^n{y}^{(n)}$ gives $r(r-1)(r-2)\cdots (r-n+1)$, so the characteristic equation is a polynomial of degree $n$ in $r$. Roots contribute the same way: $x^{r_1}$, $x^{r_1}\ln x$, $x^{r_1}(\ln x{)}^2$, and so on for multiplicities.