Cauchy-Euler Differential Equation

Subjects: Ordinary Differential Equations
Links: Second Order Euler Equation, nth Order Linear Differential Equations

The Cauchy-Euler equation is of a linear $n$ th order differential equation of the form

L [y] = \sum_{k = 0}^{n} a_{k} x^{k} y^{(k)} = 0

The way of solving it is by substituting $x = e^{u}$ , then this will help reduce it to a $n$ th order linear differential equation with constant coefficients. Alternatively, the trial solution $y = x^{m}$ may be used to directly solve for the basic solutions.

If we try the solution of the form $y = x^{r}$ , then we get a polynomial $q (r)$ , such that

L [| x^{r} |] = q (r) | x^{r} |

If we try to solve who is $q (r)$ , we get that

q (r) = \sum_{k = 0}^{n} a_{k} (r)_{n}

with $(r)_{n}$ being the falling factorial. This polynimial is called the indical polynomial of the Euler equation. Differentiating $k$ times with respect to $r$ we obtain that

\frac{\partial {#k} }{\partial r^k} L[|x^r|] = L\left[\frac{\partial {#k} }{\partial r^k} |x^r|\right] = L[x^r\ln^k |x|]

This is equal to

\frac{\partial {#k} }{\partial r^k} [q(r)|x^r|]=\left[\sum_{m = 0}^k {k\choose m}q^{(m)}(r) \ln^{k-m}|x|\right] |x|^r

if $r_{1}$ is a root with multiplicity of $m_{1}$ , then

q^{(k)} (r_{1}) = 0 for k = 0, 1, \dots, m_{1} - 1

then we know that

| x |^{r_{1}} \ln^{k} | x | for k = 0, 1, \dots, m_{1} - 1

are solutions to $L [y] = 0$ . Repeating this process for each root in $q$ we obtain that

Th: Let $r_{1}, \dots, r_{s}$ be distinct roots of the indical polynomial $q$ for an Euler-Cauchy equation, and suppose $r_{i}$ has multiplicity $m_{i}$ . Then the functions

| x |^{r_{i}} \ln^{k} | x | for k = 0, 1, \dots, m_{i} - 1

for $i = 1, \dots, s$

form a basis for the solutions of the $n$ th order Euler-Cauchy equation on any interval not contaning $x = 0$