# Suppose I am given a Cauchy-Euler form second order differential equation x^2(d^2y/dx^2)+x(dy/dx)+y=f(x).

Suppose I am given a Cauchy-Euler form second order differential equation
${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}+x\frac{dy}{dx}+y=f\left(x\right).$
The usual textbook method for solving the Cauchy Euler equation is to blackuce it to a linear differential equation with constant coefficients by the transformation $x={e}^{t}$. But I have a fundamental doubt here, we know that ${e}^{t}>0$ $\mathrm{\forall }t\in \mathbb{R}$. But when we are using the above transformation we are subconsciously assuming $x>0$. How does this make sense?

Substituting $t=\mathrm{ln}\left(x\right)$ also makes no difference as ln is defined on ${\mathbb{R}}_{>0}$. So I now doubt the validity of the method followed to solve the Cauchy-Euler equation.

Can someone give me a proper explanation of what exactly going on here and why the process is valid?
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Ali Harper
The Euler-Cauchy equation has a singularity at $x=0$. This singularity splits the domain of the ODE, which in turn limits any solution of the ODE to a sub-interval of either $x>0$ or $x<0$, depending on the initial condition. In the unusual case where the initial condition is at some negative ${x}_{0}$, you can of course also use the substitution $x=-{e}^{t}$ or more generally $x={x}_{0}{e}^{t}$.
So with $u\left(t\right)=y\left({x}_{0}{e}^{t}\right)$ one gets ${u}^{\prime }\left(t\right)={y}^{\prime }\left({x}_{0}{e}^{t}\right){x}_{0}{e}^{t}=x{y}^{\prime }\left(x\right)$ and ${u}^{″}\left(t\right)={y}^{″}\left({x}_{0}{e}^{t}\right)\left({x}_{0}{e}^{t}{\right)}^{2}+{y}^{\prime }\left({x}_{0}{e}^{t}\right){x}_{0}{e}^{t}={x}^{2}{y}^{″}\left(x\right)+x{y}^{\prime }\left(x\right)$. Here one can see that the effect of the substitution is rather independent of ${x}_{0}$ and its sign.
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Emmanuel Pace
The Eular equation can be solved y assuming $y={x}^{m}$ (Euler's substitution: ES), then we get $m=±i$, so the solution is $y={C}_{1}{x}^{i}+{C}_{2}{x}^{-i}$ as
${x}^{i}={e}^{\mathrm{ln}{x}^{i}}={e}^{i\mathrm{ln}x}=\mathrm{cos}\left(\mathrm{ln}x\right)+i\mathrm{sin}\left(\mathrm{ln}x\right).$
So the solution of (1) can also be qrutten as $y={D}_{1}\mathrm{sin}\mathrm{ln}x+{D}_{2}\mathrm{cos}\mathrm{ln}x$. So so far no problem with ES.

Alternatively one can claim that ${y}_{1}=\mathrm{sin}\mathrm{ln}x,{y}_{2}=\mathrm{cos}\mathrm{ln}x$ are two linearly independent solutions of (1). However the solution of the in-homogeneous ODE:

can be obtained by the method of variation of parameters using ${y}_{1},{y}_{2}$.