# Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE

## Theorem

Consider the Cauchy-Euler equation:

$(1): \quad x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y = 0$

By making the substitution:

$x = e^t$

it is possible to convert $(1)$ into a constant coefficient homogeneous linear second order ODE:

$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$

### General Result

Let $n \in \Z_{>0}$ be a strictly positive integer.

$a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} c + a_0 \, \map f x = 0$

can be transformed to a linear differential equation with constant coefficients by substitution $x = e^t$.

## Proof

We have:

 $\ds x$ $=$ $\ds e^t$ Derivative of Exponential Function $\ds \leadsto \ \$ $\ds \frac {\d x} {\d t}$ $=$ $\ds e^t$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds \frac {\d t} {\d x}$ $=$ $\ds \frac 1 x$ Derivative of Inverse Function $\text {(2)}: \quad$ $\ds$ $=$ $\ds e^{-t}$

Then:

 $\ds \frac {\d y} {\d x}$ $=$ $\ds \frac {\d y} {\d t} \frac {\d t} {\d x}$ Chain Rule for Derivatives $\ds$ $=$ $\ds \frac {\d y} {\d t} e^{-t}$ from $(2)$

and:

 $\ds \frac {\d^2 y} {\d x^2}$ $=$ $\ds \frac {\d} {\d x} \paren {\frac {\d y} {\d x} }$ $\ds$ $=$ $\ds \frac {\d} {\d t} \paren {\frac {\d y} {\d x} } \frac {\d t} {\d x}$ Chain Rule for Derivatives $\ds$ $=$ $\ds \frac {\d} {\d t} \paren {\frac {\d y} {\d t} e^{-t} } e^{-t}$ from $(2)$ $\ds$ $=$ $\ds e^{-t} \paren {e^{-t} \frac {\d^2 y} {\d t^2} - e^{-t} \frac {\d y} {\d t} }$ Product Rule for Derivatives $\ds$ $=$ $\ds e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} }$

Substituting back into $(1)$:

 $\ds x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds e^{2 t} \paren {e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} } } + p e^t \paren {\frac {\d y} {\d t} e^{-t} } + q y$ $=$ $\ds 0$ $\ds \frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} + p \frac {\d y} {\d t} + q y$ $=$ $\ds 0$ $\ds \frac {\d^2 y} {\d t^2} + \paren {p - 1} \frac {\d y} {\d t} + q y$ $=$ $\ds 0$

$\blacksquare$