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For the Airy ordinary differential equation $y^{\prime \prime }=xy$, a power series obeys $a_2=0$ and $a_{m+3}=a_m/[(m+3)(m+2)]$. Setting its first two coefficients to $(1,0)$ or $(0,1)$ givesEmpty products are one. The ratio test proves convergence on the whole complex plane. Termwise differentiation verifies the equation, and the initial normalization gives Wronskian one. By the Abel identity, the Wronskian remains one everywhere, so these solutions span the whole solution space.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

For real parameters, work on $x>0$ so arbitrary real powers are unambiguous. Taking derivatives givesThe residual after substitution isWhen $\gamma \ne 0$, $x^{\gamma }$ ranges over $(0,\infty )$. Evaluating at the zeros of the cosine first forces the sine coefficient to vanish; then the cosine coefficient vanishes identically. If $\gamma =0$, the nonzero constant $\cos 1$ cannot satisfy the equation. Thus the necessary and sufficient conditions areUnder these conditions the normalized coefficient is $p=(1-\gamma )/x$. The Abel identity gives $W=C{x}^{\gamma -1}$. Choose $C=\gamma$ and apply reduction of order on an interval where $\cos (x^{\gamma })\ne 0$:The substitution $u=x^{\gamma }$ evaluates the integral. After continuation across the cosine zeros,This is a power substitution for an oscillatory second-order equation: in the variable $u=x^{\gamma }$, the equation reduces to the harmonic oscillator equation.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

Here the indicial equation has the repeated root $0$. The first Frobenius method solution is $y_1=e^{-x}$. Since $p=1+1/x$, the Abel identity gives $W=C{e}^{-x}/x$. Reduction of order with $C=1$ produceswhere the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent real analytic solution does not exist.

The general local form isThis is the repeated-root Logarithmic Frobenius solution. On the negative real side one may use $\log |x|$, or use a fixed logarithm branch for a complex solution.

Assume the usual continuous-coefficient setting for a second-order linear differential equation. The Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ has derivativewhich 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 $\alpha y_1^{\prime }(x_0)+\beta y_2^{\prime }(x_0)=0$. The uniqueness theorem for ordinary differential equations makes that linear combination identically zero. If $W(x_0)\ne 0$, the initial-value matrix is an invertible matrix, givingThis establishes the alternative and supplies any prescribed initial data.

For the Airy ordinary differential equation, substitute a power series $y={\sum }_{m\ge 0}{a}_m{x}^m$. Coefficient comparison gives $a_2=0$ andChoose $(a_0,a_1)=(1,0)$ and $(0,1)$. The Airy power-series fundamental pair isEmpty 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 $A{y}_1+B{y}_2$.

Differentiating the Wronskian givesThus $W^{\prime }+pW=0$, the Abel identity.

Also,If $y_2(x_0)=0$, integration gives the reduction of order formula

For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution iswhich is a nonzero multiple of $x^{-1}$. HenceThe conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so

After division by $x$, the coefficient of $y^{\prime }$ is $P(x)=1/x-1$. The Abel identity for the Wronskian givesso

For $\lambda =1$, direct substitution confirms that $y_1=1-x$. Reduction of order gives a second solution proportional toTo extract the requested coefficients, putFor the differential operatorone findsThe analytic correction $h={\sum }_{n\ge 1}{b}_n{x}^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients giveHence

Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore