= Solution
With the indicated order of the two solutions, define the <Wronskian> by
$$
\boxed{W=y_1y_2'-y_1'y_2
=\det\begin{pmatrix}y_1&y_2\\y_1'&y_2'\end{pmatrix}.}
$$
Where $y_1\ne0$, this gives the first-order <linear differential equation>
$$
y_2'-\frac{y_1'}{y_1}y_2=\frac{W}{y_1},\qquad
\left(\frac{y_2}{y_1}\right)'=\frac{W}{y_1^2}.
$$
Integrating produces the <reduction of order> formula
$$
\boxed{y_2(x)=y_1(x)\left[C_2+\int_{x_0}^x\frac{W(s)}{y_1(s)^2}\,ds\right].}
$$
This formula is local on an interval where the known solution does not vanish; an arbitrary nonzero scaling of the <linearly independent> solution changes the normalization of $W$.
Differentiate the <Wronskian> and use the differential equation for each solution:
$$
W'=y_1y_2''-y_1''y_2
=y_1(-py_2'-qy_2)-(-py_1'-qy_1)y_2=-pW.
$$
Thus $\boxed{W'+pW=0}$, proving the <Abel identity>, and $W=C\exp(-\int p\,dx)$. For continuous coefficients on a nonsingular interval, <linearly independent> solutions have $C\ne0$.
For the particular equation, write $s=x-1$. The proposed solution is $y_1=s^2$, with $y_1'=2s$ and $y_1''=2$, so substitution gives $2s^2+2s^2-4s^2=0$. Away from the singular point $x=1$, division by $s^2$ gives $p=1/s$. The <Abel identity> therefore gives
$$
W=\frac Cs
$$
with a separate constant on each interval lying on one side of the singular point. The <reduction of order> integral is $\int W/y_1^2\,dx=C\int s^{-5}\,ds=-C/(4s^4)$, so
$$
y_2=C_2s^2-\frac C4s^{-2}.
$$
Choosing $C_2=0$, $C=-4$ gives
$$
\boxed{y_2(x)=(x-1)^{-2},\qquad W(x)=-\frac4{x-1}.}
$$
The nonzero <Wronskian> proves <linear independence>. Direct substitution also verifies the second solution, since $s^2(6s^{-4})+s(-2s^{-3})-4s^{-2}=0$. These solutions form a <basis> on either nonsingular interval; the second solution is not defined at $x=1$.
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