Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE

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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.

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

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 $(1)$:

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

$\blacksquare$


Sources

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