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

Theorem
The ordinary differential equation:


 * $$a_n x^n f^{\left({n}\right)} \left({x}\right) + \cdots + a_1 x f' \left({x}\right) + a_0 f \left({x}\right) = 0$$

can be easily transformed to linear differential equations by substitution $$x = e^t$$.

Proof
$$x=e^t$$

$$\frac{dx}{dt}=e^t=x$$

$$\frac{dt}{dx}=e^{-t}=x^{-1}$$

Base case: $$n=1$$

$$a_{1}x\frac{dy}{dx}=a_{1}e^{t}\frac{dy}{dt}\frac{dt}{dx}=a_{1}e^{t}\frac{dy}{dt}e^{-t}=a_{1}\frac{dy}{dt}$$

Induction Hypothesis:

$$a_{n}x^{n}\frac{d^{n}y}{dx^{n}}=b_{n}\frac{d^{n}y}{dt^{n}}+...+b_{1}\frac{dy}{dt}$$

$$\frac{d^{n}y}{dx^{n}}=c_{n}\frac{d^{n}y}{dt^{n}}e^{-tn}+...+c_{1}\frac{dy}{dt}e^{-tn}$$

Induction Step:  $$n=k+1$$

$$a_{n}x^{n}\frac{d^{n}y}{dx^{n}}=a_{n}e^{(k+1)t}\frac{d}{dt}\left( \frac{d^{k}y}{dx^{k}}\right) \frac{dt}{dx}=a_{n}e^{(k+1)t}\frac{d}{dt}\left( \frac{d^{k}y}{dx^{k}}\right) e^{-t}=a_{n}e^{kt}\frac{d}{dt}\left( \frac{d^{k}y}{dx^{k}}\right)=a_{n}e^{kt}\frac{d}{dt}\left( c_{k}\frac{d^{k}y}{dt^{k}}e^{-tk}+...+c_{1}\frac{dy}{dt}e^{-tk}\right)=$$

$$=a_{n}e^{kt}\left( c_{k}\frac{d^{k+1}y}{dt^{k+1}}e^{-tk}-kc_{k}\frac{d^{k}y}{dt^{k}}e^{-tk}+...+c_{1}\frac{d^{2}y}{dt^{2}}e^{-tk}-kc_{1}\frac{dy}{dt}e^{-tk}\right)=$$

$$=a_{n}c_{k}\frac{d^{k+1}y}{dt^{k+1}}-a_{n}kc_{k}\frac{d^{k}y}{dt^{k}}+...+a_{n}c_{1}\frac{d^{2}y}{dt^{2}}-a_{n}kc_{1}\frac{dy}{dt}=$$

$$=b_{n}\frac{d^{n}y}{dt^{n}}+...+b_{1}\frac{dy}{dt}$$