Solve the Cauchy - Euler equation x²y" - 4xy' - 6y = 0

The given differential equation is:

$$x^2 y'' - 4xy' - 6y = 0$$

For a Cauchy-Euler equation, we assume a solution of the form $y = x^m$.

If $y = x^m$, then:

$$y' = mx^{m-1}$$ $$y'' = m(m-1)x^{m-2}$$

Substitute the expressions for $y$, $y'$, and $y''$ into the differential equation:

$$x^2 (m(m-1)x^{m-2}) - 4x (mx^{m-1}) - 6 (x^m) = 0$$

Simplify the terms:

$$m(m-1)x^m - 4mx^m - 6x^m = 0$$

Factor out $x^m$:

$$x^m [m(m-1) - 4m - 6] = 0$$

Since $x^m \neq 0$ for $x \neq 0$, the characteristic equation is:

$$m(m-1) - 4m - 6 = 0$$ $$m^2 - m - 4m - 6 = 0$$ $$m^2 - 5m - 6 = 0$$

We can factor the quadratic equation:

$$(m-6)(m+1) = 0$$

This gives two distinct real roots:

$$m_1 = 6$$ $$m_2 = -1$$

For distinct real roots $m_1$ and $m_2$, the general solution of a Cauchy-Euler equation is given by $y(x) = c_1 x^{m_1} + c_2 x^{m_2}$.
Substituting the values of $m_1$ and $m_2$:

$$y(x) = c_1 x^6 + c_2 x^{-1}$$