= Solution
Where $y_1$ is nonzero, the <Wronskian> identity $W=y_1y_2'-y_1'y_2$ becomes the first-order equation
$$
y_2'-\frac{y_1'}{y_1}y_2=\frac W{y_1}.
$$
Its <integrating factor> is $1/y_1$, giving $(y_2/y_1)'=W/y_1^2$. Integration proves the <reduction of order> formula
$$
\boxed{y_2(x)=y_1(x)\left[C_0+\int_{x_0}^x\frac{W(t)}{y_1(t)^2}\,dt\right].}
$$
Taking $C_0=0$ gives the displayed representative in the question, with $y_2(x_0)=0$ when $y_1(x_0)\ne0$. 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:
$$
W'=y_1y_2''-y_1''y_2
=y_1(-py_2'-qy_2)-(-py_1'-qy_1)y_2=-pW.
$$
This proves <Abel's identity> \b[$W'+pW=0$] rather than merely quoting it.
For the specific equation, substituting $y_1=1-x$, $y_1'=-1$, $y_1''=0$ leaves $(1-x^2)-(1+x)(1-x)=0$. Away from zero its normalized first-derivative coefficient is $p(x)=x-1/x$, so
$$
W(x)=Cx e^{-x^2/2}.
$$
Although 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,
$$
y_2(x)=C(1-x)\int_0^x\frac{t e^{-t^2/2}}{(1-t)^2}\,dt.
$$
The integrand through degree three is
$$
\frac{t e^{-t^2/2}}{(1-t)^2}
=t\bigl(1-t^2/2+O(t^4)\bigr)\bigl(1+2t+3t^2+O(t^3)\bigr)
=t+2t^2+\frac52t^3+O(t^4).
$$
After integrating and multiplying by $1-x$,
$$
y_2(x)=C\left(\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}+O(x^5)\right).
$$
The condition $y_2''(0)=1$ fixes $C=1$, and the lower limit already gives $y_2(0)=0$. Therefore the first three nonzero terms are
$$
\boxed{y_2(x)=\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}+O(x^5).}
$$
The representation genuinely solves the second-order equation: writing $L[y]=y''+py'+qy$, differentiating its <Wronskian> gives $W'+pW=y_1L[y_2]-y_2L[y_1]=y_1L[y_2]$. Its left-hand side is zero and $y_1\ne0$ near zero. Analyticity then extends the original unnormalized equation to zero itself. Its <Wronskian> with $y_1$ is nonzero for sufficiently small $x\ne0$, 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.
Back to article page
