4.5 The Cauchy–Euler Equation

Linear equations with variable coefficients are usually hard. In general there is no formula for their solutions, and Chapter 6 will need power series to make progress. There is, however, one important family that we can solve completely with what we already know. In this family, every derivative $y^{(k)}$ is multiplied by the matching power $x^k$. A change of variable turns such an equation into one with constant coefficients, and then Sections 4.2 and 4.3 finish the job.

These equations are not a curiosity. They appear whenever a problem has radial symmetry. Steady heat conduction through the wall of a pipe gives $r^2{T}^{\prime \prime }+r{T}^{\prime }=0$. The radial displacement $u(r)$ in a thick-walled cylinder under pressure satisfies $r^2{u}^{\prime \prime }+r{u}^{\prime }-u=0$. Separating variables in Laplace's equation in polar coordinates produces $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. In each case the radius $r$ plays the role of $x$.

The equation

The name "equidimensional" comes from a scaling property. If $x$ carries a unit (say metres), then $y^{\prime }$ carries "per metre" and $y^{\prime \prime }$ carries "per metre squared". So $x^k{y}^{(k)}$ has the same units as $y$ for every $k$. Stretching the $x$-axis, $x\to \lambda x$, leaves the left-hand side unchanged in form.

The leading coefficient $a{x}^2$ vanishes at $x=0$, so the existence–uniqueness theorem of Section 4.1 does not apply on any interval containing the origin. We therefore solve on $x>0$ (and explain the case $x<0$ later).

A key observation: power functions

Why should powers of $x$ be the natural building blocks? Differentiate $x^m$ a few times:

Each term $x^k{y}^{(k)}$ turns $x^m$ into a constant times $x^m$. This is exactly how $e^{rx}$ behaves for constant-coefficient equations, where each derivative turns $e^{rx}$ into a constant times $e^{rx}$. The substitution $x=e^t$ makes the analogy precise.

The substitution x = e^t

Let $x>0$ and introduce a new independent variable $t$ by

We regard $y$ as a function of $t$ and express the $x$-derivatives through $t$-derivatives. We write $D=\frac{d}{dt}$, so $Dy=\frac{dy}{dt}$ and $D^{2}y=\frac{d^{2}y}{d{t}^2}$.

Step 1: the derivative of $t$. From $t=\ln x$,

Step 2: the first derivative. By the chain rule,

Step 3: the second derivative. Differentiate $\frac{dy}{dx}=x^{-1}\frac{dy}{dt}$ with respect to $x$, using the product rule. The factor $\frac{dy}{dt}$ is a function of $t$, so its $x$-derivative needs the chain rule again:

Multiplying by $x^2$:

Step 4: the third derivative (for later). Differentiate $\frac{d^{2}y}{d{x}^2}=x^{-2}(D^{2}y-Dy)$ in the same way:

so $x^3{y}^{\prime \prime \prime }=D^{3}y-3D^{2}y+2Dy=D(D-1)(D-2)y$.

The pattern is clear (and can be proved by induction):

Compare with the power-function computation above: the same falling products $m(m-1)\cdots (m-k+1)$ appear, with $D$ in place of $m$.

Step 5: transform the equation. Substituting $x{y}^{\prime }=Dy$ and $x^2{y}^{\prime \prime }=D^{2}y-Dy$ into $a{x}^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=g(x)$, and remembering that $x=e^t$ also on the right-hand side, gives

The shortcut: try y = x^m directly

For the homogeneous equation $a{x}^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=0$ we can skip the change of variable. Substitute $y=x^m$ and use the power-function identities:

Since $x^m\ne 0$ for $x>0$, the bracket must vanish.

The indicial polynomial is exactly the characteristic polynomial of the transformed equation in $t$ (replace $D$ by $m$). So the two routes always agree: a root $m$ gives $e^{mt}$ in the $t$-world, and $e^{mt}=(e^t{)}^m=x^m$ in the $x$-world.

The three cases for the roots

Let $m_1$, $m_2$ be the roots of the indicial equation. Translate the three cases of Section 4.2 back with $t=\ln x$:

• $e^{mt}=x^m$,
• $t{e}^{mt}=x^m\ln x$,
• $e^{\alpha t}\cos \beta t=x^{\alpha }\cos (\beta \ln x)$ and $e^{\alpha t}\sin \beta t=x^{\alpha }\sin (\beta \ln x)$.

Each pair really is a fundamental set on $x>0$, because the Wronskians never vanish there:

Solving on x < 0

On $x<0$, put $s=-x>0$. Then $\frac{dy}{dx}=-\frac{dy}{ds}$ and $\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{s}^2}$, so

The equation has exactly the same form in $s$, with $s>0$. Hence the formulas of Theorem 4.5.2 hold on $x<0$ if we replace $x$ by $|x|$ everywhere: $|x{|}^m$, $|x{|}^m\ln |x|$, $|x{|}^{\alpha }\cos (\beta \ln |x|)$, $|x{|}^{\alpha }\sin (\beta \ln |x|)$. (Equivalently, use the substitution $x=-e^t$.) The same holds for higher order, since the sign changes in $x^k{y}^{(k)}$ always pair up. Each formula is valid on one side of the origin; a solution on $x>0$ and one on $x<0$ are separate objects.

A related remark: an equation such as $a(x-x_0{)}^2{y}^{\prime \prime }+b(x-x_0)y^{\prime }+cy=0$ is a Cauchy–Euler equation in the shifted variable $x-x_0$. Try $y=(x-x_0{)}^m$.

Third and higher order

For $a_0{x}^3{y}^{\prime \prime \prime }+a_1{x}^2{y}^{\prime \prime }+a_{2}x{y}^{\prime }+a_{3}y=0$, substituting $y=x^m$ gives the cubic indicial equation

Each root contributes as in Theorem 4.5.2. A real root $m$ of multiplicity $k$ contributes $x^m,x^m\ln x,\dots ,x^m(\ln x{)}^{k-1}$ (the images of $e^{mt},t{e}^{mt},\dots ,t^{k-1}{e}^{mt}$). A complex pair contributes $x^{\alpha }\cos (\beta \ln x)$ and $x^{\alpha }\sin (\beta \ln x)$.

Nonhomogeneous equations

There are two routes for $a{x}^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=g(x)$.

Route A: transform everything. Use the transformed equation $a\ddot{y}+(b-a)\dot{y}+cy=g(e^t)$, where a dot means $d/dt$. If $g(e^t)$ is a UC function of $t$, solve it by undetermined coefficients (Section 4.3), then replace $t$ by $\ln x$. The useful translations are:

• $x^p\to e^{pt}$,
• $\ln x\to t$, and $x^p(\ln x{)}^k\to t^k{e}^{pt}$,
• $x^p\cos (\beta \ln x)\to e^{pt}\cos \beta t$, and likewise for sine.

Route B: variation of parameters in $x$. Write down $y_1,y_2$ from Theorem 4.5.2 and apply Section 4.4 directly. Remember to normalize: divide by $a{x}^2$, so that $f(x)=\frac{g(x)}{a{x}^2}$. Route B works for any continuous $g$, including those that do not become UC functions of $t$.

Try it yourself

The solver forms the indicial equation, classifies the roots, writes the complementary function and, when a right-hand side is given, finds a particular solution through the transformed equation. The first preset is $x^2{y}^{\prime \prime }-3x{y}^{\prime }-5y=9x^2$; the second is the third-order equation $x^3{y}^{\prime \prime \prime }-2x{y}^{\prime }+2y=0$, whose indicial equation has irrational roots. Enter your own coefficients to check your work.

Worked examples