Q:

General Solution of Euler's Equation

Find the general solution to the Euler's Equation:
$$
x^2\frac{d^2y}{dx^2}+2x\frac{dy}{dx}-6y=0
$$
using change of independent variable given by transformation:
$$
x = e^z
$$
Any help would be greatly appreciated.Thanks :)

A:

Hint Use the following
$$x\frac{dy}{dx}=e^z\frac{dy}{dz}*\frac{dz}{dx}
= e^z\frac{dy}{dz}*\frac{1}{e^z}=\frac{dy}{dz}$$
$$x^2\frac{d^2y}{dx^2}=e^{2z}\left(\frac{d}{dx}\frac{dy}{dx}\right)=e^{2z}\frac{d}{dx}\left(e^{-z}\frac{dy}{dz}\right)$$

A:

Another approach:
Let's look for solutions of the form $y=x^k,$ so
$$\frac{dy}{dx}=kx^{k-1}\qquad\text{and}\qquad \frac{d^2y}{dx^2}=k(k-1)x^{k-2}$$
Then
\begin{align*}
x^2\frac{d^2y}{dx^2}+2x\frac{dy}{dx}-6y=0\quad&\iff\quad x^2k(k-1)x^{k-2}+2xkx^{k-1}-6x^k=0\\
&\iff\quad x^k(k^2-k-6)=0 \quad\text{for all }x>0\\
&\iff\quad k^2-k-6=0
\end{align*}
It follows that $k=3$ and $k=-2$, then $$y=c_1x^3+c_2x^{-2}\qquad \text{for }x>0.$$

