Example 1. Find the general solution of the Euler equation $x^{2} y^{\prime \prime}+2 x y^{\prime}-$ $6 y=0$.
Solution. First method. In the equation, we make the substitution $x=e^{t}$, then

$$
\begin{aligned}
& y^{\prime}=\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=e^{-t} \frac{d y}{d t} \\
& y^{\prime \prime}=\frac{d y^{\prime}}{d x}=\frac{d y^{\prime} / d t}{d x / d t}=\frac{\left(d^{2} y / d t^{2}-d y / d t\right) e^{-t}}{e^{t}}=e^{-2 t}\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right)
\end{aligned}
$$

and the equation will take the form

$$
\frac{d^{2} y}{d t^{2}}+\frac{d y}{d t}-6 y=0
$$

The roots of the characteristic equation are $\lambda_{1}=-3, \lambda_{2}=2$, and the general solution of the equation will be $y=C_{1} e^{-3 t}+C_{2} e^{2 t}$. But since $x=e^{t}$, then $y=C_{1} x^{-3}+C_{2} x^{2}$ or

$$
y=\frac{C_{1}}{x^{3}}+C_{2} x^{2}
$$

Second method. We will look for the solution of the given equation in the form $y=x^{k}$, where $k-$ is an unknown number. We find $y^{\prime}=k x^{k-1}, y^{\prime \prime}=k(k-1) x^{k-2}$. Substituting into the equation, we get

$$
x^{2} k(k-1) x^{k-2}+2 x k x^{k-1}-6 x^{k}=0
$$

or

$$
x^{k}[k(k-1)+2 k-6]=0
$$

But since $x^{k} \not \equiv 0$, then $k(k-1)+2 k-6=0$, or $k^{2}+k-6=0$. The roots of this equation are $k_{1}=-3, k_{2}=2$. The corresponding fundamental system of solutions is $y_{1}=x^{-3}, y_{2}=x^{2}$, and the general solution will still be

$$
y=C_{1} x^{-3}+C_{2} x^{2}
$$

Non-homogeneous Euler equations of the form

$$
\sum_{k=0}^{n} a_{k} x^{k} y^{(k)}=x^{\alpha} P_{m}(\ln x)
$$

where $P_{m}(u)$ is a polynomial of degree $m$, can also be solved by trial and error, similar to solving a non-homogeneous linear differential equation with constant coefficients and a right-hand side of the form $e^{\alpha x} P_{m}(x)$.