Let $y = \map f x$. First the following are established: {{begin-eqn}} {{eqn | l = x       | r = e^t       | c = Derivative of Exponential Function }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d t}       | r = e^t       | c =  }} {{eqn | n = 1       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d t} {\d x}       | r = \frac 1 x       | c = Derivative of Inverse Function }} {{eqn | n = 2       | r = e^{-t}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = a_1 x \frac {\d y} {\d x}       | r = a_1 e^t \frac {\d y} {\d t} \frac {\d t} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = a_1 e^t \frac {\d y} {\d t} e^{-t}       | c = from $(2)$ }} {{eqn | r = a_1 \frac {\d y} {\d t}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = a_k x^k \dfrac {\d^k y} {\d x^k}       | r = 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}       | c =  }} {{eqn | l = \frac {\d^k y} {\d x^k}       | r = 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}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = a_{k + 1} x^{k + 1} \dfrac {\d^{k + 1} y} {\d x^{k + 1} }       | r = 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}       | c =  }} {{eqn | l = \frac {\d^{k + 1} y} {\d x^{k + 1} }       | r = 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} }       | c =  }} {{end-eqn}} === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = a_{k + 1} x^{k + 1} \frac {\d^{k + 1} y} {\d x^{k + 1} }       | r = 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} }} {{eqn | r = a_{k + 1} e^{\paren {k + 1} t} \, \map {\frac \d {\d t} } {\frac {\d^k y} {\d x^k} } e^{-t} }} {{eqn | r = a_{k + 1} e^{k t} \, \map {\frac \d {\d t} } {\frac {\d^k y} {\d x^k} } }} {{eqn | r = 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} } }} {{eqn | r = 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} } }} {{eqn | r = 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} }} {{eqn | r = 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} }} {{end-eqn}} Hence the result by the Principle of Mathematical Induction. {{qed}} Category:Cauchy-Euler Equation eeo6xd6tfoaxnucn3qjwq9pww5lnsy2