First, you can write the equation in the form L(x)y(x)=0 where
L(x) = x^3\frac{d^2}{dx^2}+x^2\frac{d}{dx}-x(x^2+1)
is a linear operator acting on y(x). What you've figured out is that L(x) = -L(-x). In other words, L(x) has a definite parity, and it is odd. When you refer to the parity of an equation, you're just talking about the parity of the operator part.
Second, if you change variables from x to -x, you have
L(x)y(x)=0 → L(-x)y(-x) = 0 → -L(x)y(-x) = 0 → L(x)y(-x) = 0
This means if y(x) is a solution, so is y(-x). Note we're not saying anything about the parity of y(x). The functions y(x) and y(-x) could be linearly independent; all we know is that they're both solutions to the differential equation.
Finally, since L(x) is a linear operator, any linear combination of solutions is also a solution to the differential equation. In particular, you can form the combinations
\begin{align*}<br />
y_e(x) &= y(x) + y(-x) \\<br />
y_o(x) &= y(x) - y(-x)<br />
\end{align*}<br />
and they have a definite parity.
For example, suppose you have the differential equation
y'' + y = \left[\frac{d^2}{dx^2}+1\right]y(x) = 0
which has even parity. One solution to the equation is y1(x) = eix. Because the equation has a definite parity, we know that y2(x) = e-ix is also a solution. Neither of these solutions is even or odd, but you can always find ones that are:
\begin{align*}<br />
y_e(x) &= e^{ix} + e^{-ix} = 2\cos x \\<br />
y_o(x) &= e^{ix} - e^{-ix} = 2i\sin x<br />
\end{align*}<br />
