= Solution
Assume the usual continuous-coefficient setting for a <second-order linear differential equation>. The <Wronskian> $W=y_1y_2'-y_1'y_2$ has derivative
$$
W'=y_1y_2''-y_1''y_2=-p(x)W,
\qquad
W(x)=W(x_0)\exp\!\left[-\int_{x_0}^xp(s)\,ds\right],
$$
which proves the <Abel identity>. If $W(x_0)=0$, the two initial-value columns are <linearly dependent>, so some nonzero $(\alpha,\beta)$ gives $\alpha y_1(x_0)+\beta y_2(x_0)=0$ and $\alpha y_1'(x_0)+\beta y_2'(x_0)=0$. The <uniqueness theorem for ordinary differential equations> makes that linear combination identically zero. If $W(x_0)\ne0$, the initial-value matrix is an <invertible matrix>, giving
$$
\boxed{a=\frac{Ay_2'(x_0)-By_2(x_0)}{W(x_0)},\qquad
b=\frac{By_1(x_0)-Ay_1'(x_0)}{W(x_0)}.}
$$
This establishes the alternative and supplies any prescribed initial data.
For the <Airy ordinary differential equation>, substitute a <power series> $y=\sum_{m\geq0}a_mx^m$. Coefficient comparison gives $a_2=0$ and
$$
a_{m+3}=\frac{a_m}{(m+3)(m+2)},\qquad m\geq0.
$$
Choose $(a_0,a_1)=(1,0)$ and $(0,1)$. The <Airy power-series fundamental pair> is
$$
\boxed{y_1(x)=\sum_{k=0}^\infty
\frac{x^{3k}}{\prod_{j=1}^k(3j)(3j-1)},\qquad
y_2(x)=\sum_{k=0}^\infty
\frac{x^{3k+1}}{\prod_{j=1}^k(3j)(3j+1)}.}
$$
Empty products mean one. Their first terms are $y_1=1+x^3/6+x^6/180+\cdots$ and $y_2=x+x^4/12+x^7/504+\cdots$. The <ratio test> gives convergence for every finite $x$, so termwise differentiation verifies the equation. Their <Wronskian> is one at zero, and hence everywhere because $p=0$. Thus every solution is $Ay_1+By_2$.
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