 Solution

On an interval where $p,q$ are continuous, define the WronskianUsing the two second-order linear differential equations to eliminate the second derivatives givesThe integrating factor yields the Abel identityThus 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 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 $y^{\prime }$ is zero, so its Wronskian is constant there. The proposed $y_1=x^2$ satisfies $x^2{y}_1^{\prime \prime }-2y_1=2x^2-2x^2=0$. To construct a second solution by reduction of order, imposeDividing by $x^4$ shows that $(y_2/x^2{)}^{\prime }=C/x^4$, givingTaking $C=-3,D=0$ producesThe nonzero Wronskian proves linear independence; hence the general solution on either nonsingular interval is $y=A{x}^2+B/x$. Zero is a singular point of a differential equation and is excluded from this construction.