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:
\(\ds x\) | \(=\) | \(\ds e^t\) | 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.
$\blacksquare$