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For $y\prime \prime +py\prime +qy=0$, the Wronskian satisfies $W\prime =-pW$.

More precisely, let $y_1,y_2$ be solutions of a differential equation for the same homogeneous linear differential equationon an interval $I$ where $p,q$ are continuous functions. For any $x_0\in I$, Abel's identity givesOnly the coefficient of $y^{\prime }$ enters the evolution of the Wronskian. In an unnormalized equation $A(x)y^{\prime \prime }+B(x)y^{\prime }+C(x)y=0$, first use $p=B/A$ on an interval where $A\ne 0$ and the coefficient ratios are continuous functions.^[ref]

The proof uses cancellation. Differentiating the Wronskian by the product rule and substituting the common equation givesMultiplying this first-order linear differential equation by the integrating factor $\mu (x)=\exp ({\int }_{x_0}^{x}p(s)ds)$ gives $(\mu W{)}^{\prime }=0$. Since $\mu (x_0)=1$, the constant is $W(x_0)$, proving Abel's identity. This argument also covers $W(x_0)=0$ without dividing by $W$.

The exponential function is positive for real $p$, so a real Wronskian either vanishes everywhere or has a fixed nonzero sign throughout $I$. In the preceding example, $p=-1$ and $W(0)=3$ give $W(x)=3e^x$ immediately. If $p=0$, every Wronskian of two solutions of a differential equation for that same equation is constant, regardless of $q$. Thus Abel's identity can certify linear independence and find the denominator used in variation of parameters without explicit formulas for both solutions of a differential equation.

Named after George Biddell Airy, the standard Airy functions solve the Airy ordinary differential equation. Their Wronskian is $\mathrm{Ai}(Z){\mathrm{Bi}}^{\prime }(Z)-{\mathrm{Ai}}^{\prime }(Z)\mathrm{Bi}(Z)=1/\pi$, so they form a fundamental solution pair, including for complex $Z$.

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.

A regular singular point with a full set of analytic local solutions is an apparent singularity of the differential equation. Singular normalized coefficients do not by themselves force singular solutions. For example, $x{y}^{\prime \prime }-y^{\prime }=0$ has normalized coefficient $-1/x$, but its fundamental solutions $1,x^2$ are both analytic through zero. Their Wronskian is $2x$, which vanishes at the singular point while being nonzero on each adjacent ordinary interval. The nonvanishing-Wronskian criterion for a fundamental set applies on intervals where the normalized equation has regular coefficients.

This second-order linear differential equation extends the Legendre differential equation by its order parameter $m$. Its solutions are associated Legendre functions. In a degree-one potential problem, with $x=\tanh \eta$, two explicit solutions are $e^{m\eta }(m-\tanh \eta )$ and $e^{-m\eta }(m+\tanh \eta )$. Their Wronskian with respect to $\eta$ is $2m(1-m^2)$, so the two formulas are not a fundamental pair at $m=0$ or $m=\pm 1$.

For $L=d^2/d{x}^2-2/x^2$ on $x>0$, the homogeneous solutions $x^2,x^{-1}$ satisfy respectively the zero and infinity boundary conditions. Their Wronskian is $-3$. The displayed Green's function is continuous at $x=\xi$ and its first derivative has jump one, so $L_{x}G=\delta (x-\xi )$ in the distribution sense. Integration against a compactly supported forcing gives the unique solution decaying at both boundaries.

A bounded function on $[0,\infty )$ has a Laplace transform satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its Laplace transform cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an ordinary differential equation into a first-order differential equation for its Laplace transform.

For example, transforming $t{y}^{\prime \prime }+(t-1)y^{\prime }-y=0$ with $y(0)=1$ gives $p(p+1)Y^{\prime }+(3p+2)Y=2$. An integrating factor $p^2(p+1)$ yieldsThe bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero Wronskian $t{e}^{-t}$ on $(0,\infty )$, so they form a fundamental system of solutions. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.

For a minimally coupled massless scalar field in flat de Sitter spacetime, $a=-1/(H\tau )$ and the Canonically rescaled de Sitter scalar mode is $u_k=a{f}_k$. The equation $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k=0$ and the positive-frequency condition $u_k\sim e^{-ik\tau }/\sqrt{2k\mathcal{V}}$ select the displayed mode. Direct differentiation verifies its Wronskian normalization, which ensures the equal-time canonical commutation relations with box-normalized creation and annihilation operators. A conformally coupled massless scalar has a different mode equation.

For $L=y^{\prime \prime }+\alpha y^{\prime }/x+(\alpha -1{)}^{2}y/(4x^2)$ on $x>0$, the repeated indicial root is $r=(1-\alpha )/2$ and the homogeneous solutions are $x^r,x^r\log x$. Their Wronskian is $x^{-\alpha }$. The displayed causal Green function is continuous at each source $\xi >0$, has a unit jump in its first derivative, and satisfies $L_{x}G=\delta (x-\xi )$. It is zero near the origin, so its value and first derivative there vanish. Those origin conditions alone fail to give uniqueness when $\alpha <-1$: then both homogeneous solutions and their first derivatives vanish as $x\downarrow 0$. For forcing $f(x)=x$, source integration from zero converges when $\alpha >-5$ and gives $y_p=4x^3/(\alpha +5{)}^2$. All zero-origin-data solutions in $-5<\alpha <-1$ are $y_p+x^r(C_1+C_2\log x)$; for $\alpha \ge -1$ the solution is unique. Causality selects the displayed kernel even when the bare singular-endpoint conditions are insufficient.

For the gap $a<r<b$, put $u_{\ell }(r)=r^{\ell }-a^{2\ell +1}{r}^{-\ell -1}$, $v_{\ell }(r)=r^{-\ell -1}-r^{\ell }/b^{2\ell +1}$ and $d_{\ell }=(2\ell +1)[1-(a/b{)}^{2\ell +1}]$. With orthonormal spherical harmonics, the Dirichlet Green function isFor a short verification, $u_{\ell }(a)=v_{\ell }(b)=0$ and both solve the radial Laplace equation. Their Wronskian is $u_{\ell }{v}_{\ell }^{\prime }-u_{\ell }^{\prime }{v}_{\ell }=-d_{\ell }/r^2$. Thus the radial derivative jump is $-1/r^{\prime 2}$; completeness of the spherical harmonics gives the stated Dirac delta function normalization. A point charge has electric potential $qG/(4\pi {ϵ}_0)$, and the inner-wall surface charge density is $-q{\partial }_{r}G{|}_{a^{+}}/(4\pi )$, with the angular source coordinates held fixed.

Let $0<a<b$ and impose zero Dirichlet boundary conditions at $a,b$. With the convention $L_{x}G(x,s)=\delta (x-s)$ for the Dirac delta distribution, the Green's function is continuous and its first derivative has jump one. For $n\ne 0$ choose $\ell (x)=x^{-n}-a^{-n}$ and $r(x)=x^{-n}-b^{-n}$. Their Wronskian is $n(a^{-n}-b^{-n})s^{-n-1}$, soFor $n=0$, take the continuous limit or use $\ell =\log (x/a)$ and $r=\log (x/b)$ to obtainThe solution is $u(x)={\int }_a^{b}G(x,s)f(s)ds$. These kernels use ordinary measure $ds$. The factor $s^{n+1}$ comes from the nonconstant Wronskian and must not be discarded by confusing the original operator with its multiplied divergence form $(x^{n+1}{u}^{\prime }{)}^{\prime }$.

For $L=d^2/d{x}^2+4x^{-1}d/dx-4x^{-2}$ on $(0,1)$, regularity at zero selects the homogeneous solution $u=x$, while the right Dirichlet boundary condition selects $v=x^{-4}-x$. Their Wronskian is $-5x^{-4}$. Thus the displayed Green function, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$, is continuous, has unit first-derivative jump and satisfies $L_{x}G=\delta (x-\xi )$. For bounded forcing $f$, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ is the unique solution continuous at zero with zero endpoint values. The singular homogeneous branch $x^{-4}$ is excluded by that regularity. The kernel uses ordinary source measure $d\xi$, not the measure for the multiplied self-adjoint operator.

For ${\partial }_x^2+k^2$ on $(0,\ell )$ with zero endpoint values and $\sin (k\ell )\ne 0$, let $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function is continuous at $x=\xi$ and its first derivative has jump one, giving unit Dirac delta forcing. The formula follows from the left solution $\sin (kx)$ and right solution $\sin (k(\ell -x))$, whose Wronskian is $-k\sin (k\ell )$. At $k=0$ its limit is $-x_{<}(\ell -x_{>})/\ell$, although zero is excluded by a condition that $k$ is not an integer in the length-$\pi$ version.

For $\kappa >0$, the Green function of $D_x^2-{\kappa }^2$ on $(0,L)$ with zero Dirichlet boundary conditions is the displayed negative kernel. Here $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The left and right homogeneous solutions are $\sinh (\kappa x)$ and $\sinh (\kappa (L-x))$, with Wronskian $-\kappa \sinh (\kappa L)$. The kernel is continuous and its first derivative jumps by $+1$, giving $(D_x^2-{\kappa }^2)G=\delta (x-\xi )$. Reversing the operator sign reverses the kernel sign.

For two solutions of the linear recurrence relation $x_{n+2}+b_n{x}_{n+1}+c_n{x}_n=0$, their discrete Wronskian obeys $W_{n+1}=c_n{W}_n$. Substitute the recurrence into $p_{n+1}{q}_{n+2}-p_{n+2}{q}_{n+1}$: the $b_n$ terms cancel and the remaining expression is $c_n(p_n{q}_{n+1}-p_{n+1}{q}_n)$. Iteration gives $W_{n+1}=W_1{\prod }_{m=1}^n{c}_m$, including when some $c_m$ vanish. A nonzero discrete Wronskian proves linear independence of the two solutions.

This nonlinear ordinary differential equation governs the amplitude of a complex oscillator solution. Independent real solutions of $y^{\prime \prime }+p(x)y=0$ can be combined into a nonvanishing complex solution; its modulus satisfies the equation with $c$ equal to the squared Wronskian. A real positive amplitude and a conserved phase flux give an exact representation, not merely a WKB approximation.

For $y_{tt}+e^{-2\epsilon t}(\epsilon y_t+y)=0$, $y(0)=0$, $y_t(0)=1$, set $r=e^{-2\epsilon t}/2$. The exact equation becomes $r{y}_{rr}+(1-r)y_r+y/(2{\epsilon }^2)=0$. Let $m(t)=M(-1/(2{\epsilon }^2),1,r)$, using the Kummer function. It is regular at $r=0$ and tends to one. The Wronskian $W=m{y}_t-m_{t}y$ obeys $W_t=-\epsilon e^{-2\epsilon t}W$, with $W(0)=M(-1/(2{\epsilon }^2),1,1/2)$. Integration gives the displayed exact late slope. The other local solution grows at most logarithmically in $r$, so $m_{t}y\to 0$ and $W\to y_t$. This provides a check on a WKB approximation even near phases where its leading predicted slope vanishes.

Suppose $y^{\prime \prime }-P(x)y^{\prime }/x-Q(x)y/x=0$ has analytic $P,Q$ near zero, $P(x)=P_0+P_{1}x+\cdots$, and a solution $y_1=1+\beta x+\cdots$, with $P_0$ a positive integer. The indicial roots are $0,P_0+1$. The Abel identity gives $W=C{x}^{P_0}(1+P_{1}x+\cdots )$. Dividing by $y_1^2$ gives $C{x}^{P_0}[1+(P_1-2\beta )x+\cdots ]$. Integrating this reduction of order expression from zero, multiplying by $y_1$, and choosing $C=P_0+1$ proves the displayed normalized expansion. The integral's zero constant eliminates the added multiple of $y_1$. Its nonzero Wronskian proves independence. In this setting the assumed exponent-zero solution prevents a logarithm in the constructed exponent-$P_0+1$ solution.

A fundamental system for a homogeneous linear ordinary differential equation of order $n$ on an interval is a basis of its vector space of homogeneous solutions. Every solution has a unique expression ${\sum }_{j=1}^n{c}_j{y}_j$ with constant coefficients. Under the usual continuity assumptions and a nonvanishing leading coefficient, a nonzero Wronskian at one point proves that $n$ solutions form such a system: their initial-value vectors are independent and span all possible initial data, and the Picard-Lindelöf theorem supplies uniqueness. For a first-order vector system, the corresponding columns form a fundamental matrix.

The homogeneous Cauchy-Euler equation $Ly=0$ has fundamental solutions $x,x^{-1}$ on $x>0$, with Wronskian $-2/x$. Requiring decay at both endpoints gives the unweighted Green's functionRequiring zero initial value and derivative at zero instead gives the forward kernelBoth kernels are continuous and have first-derivative jump one at $x=\xi$, so $L_{x}G=\delta (x-\xi )$. Away from that point they solve the homogeneous equation. Integrating against the forcing with ordinary measure $d\xi$ yields the respective solutions whenever the integral and endpoint limits converge. Their difference is $G_0-G_D=x/2$, independent of $\xi$, so the solutions differ by the homogeneous term $(x/2)\int f(\xi )d\xi$ for integrable forcing. For example, the two choices for an indicator forcing on $(0,1)$ differ by $x/2$. Specifying a differential operator without its endpoint conditions does not specify its Green kernel.