 Solution

Where $y_1$ is nonzero, the Wronskian identity $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ becomes the first-order equationIts integrating factor is $1/y_1$, giving $(y_2/y_1{)}^{\prime }=W/y_1^2$. Integration proves the reduction of order formulaTaking $C_0=0$ gives the displayed representative in the question, with $y_2(x_0)=0$ when $y_1(x_0)\ne 0$. Changing the lower limit adds a multiple of $y_1$ and does not change the linearly independent solution modulo the known one. The scale of $W$ separately fixes the normalization of $y_2$.

Differentiate the determinant and use the original second-order equation for each solution:This proves Abel's identity $W^{\prime }+pW=0$ rather than merely quoting it.

For the specific equation, substituting $y_1=1-x$, $y_1^{\prime }=-1$, $y_1^{\prime \prime }=0$ leaves $(1-x^2)-(1+x)(1-x)=0$. Away from zero its normalized first-derivative coefficient is $p(x)=x-1/x$, soAlthough the equation is singular at zero, this expression extends analytically there. Near zero, $y_1$ is nonzero and the reduction of order integrand is regular. With the prescribed lower limit,The integrand through degree three isAfter integrating and multiplying by $1-x$,The condition $y_2^{\prime \prime }(0)=1$ fixes $C=1$, and the lower limit already gives $y_2(0)=0$. Therefore the first three nonzero terms areThe representation genuinely solves the second-order equation: writing $L[y]=y^{\prime \prime }+p{y}^{\prime }+qy$, differentiating its Wronskian gives $W^{\prime }+pW=y_{1}L[y_2]-y_{2}L[y_1]=y_{1}L[y_2]$. Its left-hand side is zero and $y_1\ne 0$ near zero. Analyticity then extends the original unnormalized equation to zero itself. Its Wronskian with $y_1$ is nonzero for sufficiently small $x\ne 0$, proving linear independence. The vanishing Wronskian at the singular point zero does not contradict linear independence on a regular interval. The integral representation is used locally near zero, before the zero of $y_1$ at one.