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

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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:

\(\ds x\) \(=\) \(\ds e^t\) 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
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds e^{-t}\)


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:

\(\ds a_1 x \frac {\d y} {\d x}\) \(=\) \(\ds a_1 e^t \frac {\d y} {\d t} \frac {\d t} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds a_1 e^t \frac {\d y} {\d t} e^{-t}\) from $(2)$
\(\ds \) \(=\) \(\ds a_1 \frac {\d y} {\d t}\)

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:

\(\ds a_k x^k \dfrac {\d^k y} {\d x^k}\) \(=\) \(\ds b_k \frac {\d^k y} {\d t^k} + b_{k - 1} \frac {\d^{k - 1} y} {\d t^{k - 1} } + \dotsb + b_1 \frac {\d y} {\d t}\)
\(\ds \frac {\d^k y} {\d x^k}\) \(=\) \(\ds c_k \frac {\d^k y} {\d t^k} e^{-t k} + c_{k - 1} \frac {\d^{k - 1} y} {\d t^{k - 1} } e^{-t n} + \dotsb + c_1 \frac {\d y} {\d t} e^{-t k}\)


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:

\(\ds a_{k + 1} x^{k + 1} \dfrac {\d^{k + 1} y} {\d x^{k + 1} }\) \(=\) \(\ds b_{k + 1} \frac {\d^{k + 1} y} {\d t^{k + 1} } + b_k \frac {\d^k y} {\d t^k} + \dotsb + b_1 \frac {\d y} {\d t}\)
\(\ds \frac {\d^{k + 1} y} {\d x^{k + 1} }\) \(=\) \(\ds c_{k + 1} \frac {\d^{k + 1} y} {\d t^{k + 1} } e^{-t \paren {k + 1} } + c_k \frac {\d^k y} {\d t^k} e^{-t \paren {k + 1} } + \dotsb + c_1 \frac {\d y} {\d t} e^{-t \paren {k + 1} }\)


Induction Step

This is our induction step:


\(\ds a_{k + 1} x^{k + 1} \frac {\d^{k + 1} y} {\d x^{k + 1} }\) \(=\) \(\ds a_{k + 1} e^{\paren {k + 1} t} \, \map {\frac \d {\d t} } {\frac {\d^k y} {\d x^k} } \frac {\d t} {\d x}\)
\(\ds \) \(=\) \(\ds a_{k + 1} e^{\paren {k + 1} t} \, \map {\frac \d {\d t} } {\frac {\d^k y} {\d x^k} } e^{-t}\)
\(\ds \) \(=\) \(\ds a_{k + 1} e^{k t} \, \map {\frac \d {\d t} } {\frac {\d^k y} {\d x^k} }\)
\(\ds \) \(=\) \(\ds a_{k + 1} e^{k t} \, \map {\frac \d {\d t} } {c_k \frac {\d^k y} {\d t^k} e^{-t k} + c_{k - 1} \frac {\d^{k - 1} y} {\d t^{k - 1} } e^{-t k} + \dotsb + c_1 \frac {\d y} {\d t} e^{-t k} }\)
\(\ds \) \(=\) \(\ds a_{k + 1} e^{k t} \paren {c_k \frac {\d^{k + 1} y} {\d t^{k + 1} } e^{-t k} - k c_k \frac {\d^k y} {\d t^k} e^{-t k} + c_{k - 1} \frac {\d^k y} {\d t^k}e^{-t k} - k c_{k - 1} \frac {\d^{k - 1} y} {\d t^{k - 1} } e^{-t k} + \dotsb + c_1 \frac {\d^2 y} {\d t^2} e^{-t k} - k c_1 \frac {\d y} {\d t} e^{-t k} }\)
\(\ds \) \(=\) \(\ds a_{k + 1} c_k \frac {\d^{k + 1} y} {\d t^{k + 1} } - a_n k c_k \frac {\d^k y} {\d t^k} + c_{k - 1} \frac {\d^k y} {\d t^k} - k c_{k - 1} \frac {\d^{k - 1} y} {\d t^{k - 1} } + \dotsb + a_n c_1 \frac {\d^2 y} {\d t^2} - a_n k c_1 \frac {\d y} {\d t}\)
\(\ds \) \(=\) \(\ds b_{k + 1} \frac {\d^{k + 1} y} {\d t^{k + 1} } + b_k \frac {\d^k y} {\d t^k} + \dotsb + b_1 \frac {\d y} {\d t}\)

Hence the result by the Principle of Mathematical Induction.

$\blacksquare$