Euler Equations

Main idea

In $ax^2y'' + bxy' + cy = 0$ , each derivative comes paired with a matching power of $x$ , so $y = x^r$ turns every term into a multiple of $x^r$ . That leaves a quadratic for $r$ , $ar(r - 1) + br + c = 0$ , which plays the role of the characteristic equation.

ax^2y'' + bxy' + cy = 0 \implies a{\htmlData{sym=0}{r}}({\htmlData{sym=0}{r}} - 1) + b{\htmlData{sym=0}{r}} + c = 0

What the symbols mean

$r$
Exponent
Substituting y = xʳ turns each x-power times derivative into a multiple of xʳ.

Explanation

An Euler equation is simply a second order equation where each derivative is multiplied by a matching power of $x$ , which makes a plain power $x^r$ solve it. (It's named after the mathematician Leonhard Euler, and you'll also see it called a Cauchy-Euler equation.) It shows up wherever the physics looks the same at every scale, like stress around a hole or in a thick pipe. The equation $ax^2y'' + bxy' + cy = 0$ looks heavy, but it's just three matched pairs: $x^2y''$ (the second derivative, with two powers of $x$ to make up for the two it lost), $xy'$ (the first derivative, with one power of $x$ to make up for the one it lost) and $y$ (on its own). Because every pair has the same balance, trying $y = x^r$ turns the whole thing into an ordinary quadratic for $r$ . It looks complex, but it's the characteristic equation in disguise; bear with it.

Take a look at $ax^2y'' + bxy' + cy = 0$ . Its coefficients aren't constant, and at $x = 0$ the leading coefficient $ax^2$ vanishes, which is exactly the kind of singular point where the last lesson's series can run into trouble. But the equation has a special pattern: $y''$ comes with $x^2$ , $y'$ comes with $x$ , and $y$ comes with nothing.

Think about why that pattern helps. Each derivative of $x^r$ lowers the power by one, and the matching power of $x$ puts it right back: $xy' = rx^r$ and $x^2y'' = r(r - 1)x^r$ . So substituting $y = x^r$ (for $x > 0$ ) makes every term a multiple of $x^r$ , and dividing it out leaves $ar(r - 1) + br + c = 0$ . It's the same move as $e^{rt}$ for constant coefficients, and the three cases mirror those equations:

Roots	Solutions
real, distinct $r_1, r_2$	$x^{r_1}$ , $x^{r_2}$
repeated $r$	$x^r$ , $x^r\ln x$
complex $\lambda \pm \mu i$	$x^\lambda\cos(\mu\ln x)$ , $x^\lambda\sin(\mu\ln x)$

Why does the table look so familiar? The substitution $x = e^t$ turns an Euler equation into a constant-coefficient one. Under it, $x^r$ becomes $e^{rt}$ , and $\ln x$ becomes $t$ . So the extra $t$ for a repeated root shows up as $\ln x$ , and $\cos\mu t$ shows up as $\cos(\mu\ln x)$ .

A common mistake. It's tempting to write $ar^2 + br + c = 0$ , because the coefficients $a$ , $b$ , $c$ look just like a constant-coefficient equation's. It isn't right: $x^2y''$ contributes $r(r - 1)$ , not $r^2$ . Always substitute $x^r$ and work out the powers, or use $ar(r - 1) + br + c = 0$ directly.

See it

Stress in a thick pipe leads to the Euler equation r²u″ + ru′ − u = 0, whose power solutions are r and 1/r. The actual displacement through the wall (1 ≤ r ≤ 3) is a mix of the two, with the amounts set by the pressures inside and outside. Notice that 1/r dominates near the inner wall and r near the outer one, so the displacement dips to its smallest value, 2, at r = 2.

Worked examples

Example 1

Solve $2x^2y'' + 3xy' - 15y = 0$ .

1
Substituting $y = x^r$ gives $2r(r - 1) + 3r - 15 = 2r^2 + r - 15 = 0$ , so $r = \frac52$ or $r = -3$ .

$y = c_1x^{5/2} + c_2x^{-3}$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the roots for $2x^2y'' + 3xy' - 15y = 0$ .

Problem 2+5 XP

Solve $x^2y'' - 2y = 0$ , $y(1) = 1$ , $y'(1) = 1$ .

Problem 3+5 XP

Which pair solves $x^2y'' + xy' + 4y = 0$ ?

Problem 4+5 XP

Give a second solution of $x^2y'' - xy' + y = 0$ besides $y = x$ .

Problem 5+5 XP

Find the repeated root of the indicial equation for $x^2y'' + 5xy' + 4y = 0$ .

Problem 6+5 XP

The radial displacement $u$ (in mm) in a thick pipe wall satisfies $r^2u'' + ru' - u = 0$ for $1 \le r \le 2$ (in cm). Both surfaces move $0.3$ mm: $u(1) = 0.3$ and $u(2) = 0.3$ . Find $u(r)$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $x^2y'' + 3xy' + 5y = 0$ with $y(1) = 2$ and $y'(1) = 0$ , for $x > 0$ .

Where it's used

Euler equations describe stress around a hole in a plate and fields near sharp corners, where the scale-invariant $x^r$ behavior matters.

Stress in a Thick PipeDiff Eq · Physics

Series Solutions Higher Order Linear Equations