Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE/General Result

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

The ordinary differential equation:


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

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

Proof
Let $y = \map f x$.

First the following are established:

The proof now proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $a_n x^n \, \map {f^{\paren n} } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$

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

Basis for the Induction
$\map P 1$ is the case:

Thus:
 * $a_1 x \, \map {f'} x + a_0 \, \map f x = 0$

has been transformed into:
 * $a_1 \dfrac \d {\d t} \map f {e^t} + a_0 \, \map f {e^t} = 0$

which is a linear first Order ODE with constant coefficients.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $a_k x^k \, \map {f^{\paren k} } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$

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

Then we need to show that:
 * $a_{k + 1} x^{k + 1} \map {f^{\paren {k + 1} } } x + \dotsb + a_1 x \, \map {f'} x + a_0 \, \map f x = 0$

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

Induction Step
This is our induction step:

Hence the result by the Principle of Mathematical Induction.