= Solution
On an interval where $p,q$ are continuous, define the <Wronskian>
$$
W[u_1,u_2]=\det\begin{pmatrix}u_1&u_2\\u_1'&u_2'\end{pmatrix}
=u_1u_2'-u_1'u_2.
$$
Using the two <second-order linear differential equations> to eliminate the second derivatives gives
$$
W'=u_1u_2''-u_1''u_2=-pW.
$$
The <integrating factor> yields the <Abel identity>
$$
\boxed{W(x)=W(x_0)\exp\left(-\int_{x_0}^x p(s)\,ds\right).}
$$
Thus the <Wronskian> either vanishes identically or is nonzero everywhere on the interval. If $u_1,u_2$ are <linearly dependent>, it is identically zero. Conversely, if $W(x_0)=0$, the two initial-data vectors are dependent, so some nontrivial linear combination has both value and first derivative zero at $x_0$. The <uniqueness theorem for ordinary differential equations> makes that combination identically zero. Hence \b[the solutions are linearly independent exactly when their Wronskian is nonzero at one, and therefore every, point]. Continuity and the common differential equation are essential to this converse.
For the particular equation, divide by $x^2$ on either interval $x>0$ or $x<0$. The coefficient of $y'$ is zero, so its <Wronskian> is constant there. The proposed $y_1=x^2$ satisfies $x^2y_1''-2y_1=2x^2-2x^2=0$. To construct a second solution by <reduction of order>, impose
$$
x^2y_2'-2xy_2=C\ne0.
$$
Dividing by $x^4$ shows that $(y_2/x^2)'=C/x^4$, giving
$$
y_2=Dx^2-\frac{C}{3x}.
$$
Taking $C=-3,D=0$ produces
$$
\boxed{y_2(x)=x^{-1},\qquad W[x^2,x^{-1}]=-3.}
$$
The nonzero <Wronskian> proves <linear independence>; hence the <general solution> on either nonsingular interval is $y=Ax^2+B/x$. Zero is a <singular point of a differential equation> and is excluded from this construction.
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