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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.

variation of parameters with the fundamental solutions $x,x^2$ givesThe kernel is continuous across $x=t$ and its first derivative jumps by $t^{-2}$, which gives unit in a ring source after multiplication by the leading coefficient. It satisfies both endpoint conditions. The homogeneous solution $Ax+B{x}^2$ satisfies them only when $A=B=0$, proving uniqueness.

For $y=\alpha x+\beta x^2$, the derivatives are $y^{\prime }=\alpha +2\beta x$ and $y^{\prime \prime }=2\beta$. Substituting gives $2\beta x^2-2x(\alpha +2\beta x)+2(\alpha x+\beta x^2)=0$. The independent solutions $x,x^2$ have Wronskian $x^2\ne 0$ on $[1,2]$, so they form a fundamental system of solutions.

Divide the inhomogeneous linear differential equation by $x^2$ and use variation of parameters. Choosing integrals from one gives the particular solutionThus $z=Ax+B{x}^2+z_p$. The first boundary condition gives $A=-B$. The integral endpoint terms cancel when differentiating $z_p$, so the second givesConsequentlyCollect the coefficient of $f(t)$ on either side of $t=x$. A suitable Green function isThe two pieces agree at $x=t$, $G(1,t)=0$, and $G_x(2,t)=0$ for $t<2$. For an interior source point its derivative jump isThus multiplication by the leading coefficient $t^2$ gives unit in a ring source strength. Away from $x=t$, both pieces solve the homogeneous linear differential equation. With continuous $f$, the integral representation has the required second derivative and solves the equation pointwise in $(1,2)$.

For uniqueness, the difference of two solutions is $Ax+B{x}^2$. Its boundary conditions give $A+B=0$ and $A+4B=0$, forcing $A=B=0$. Hence the solution is unique. This is the mixed Green function for an Euler operator with exponents one and two.

If $Au+Bv=0$, evaluation at zero gives $A=0$ and differentiation at zero gives $B=0$. Thus the solutions are linearly independent. For $w=u{v}^{\prime }-u^{\prime }v$,so $\boxed{w(x)=\exp (-{\int }_0^{x}p(s)ds)>0}$.

Variation of parameters sets $y=A(x)u+B(x)v$ with $A^{\prime }u+B^{\prime }v=0$, $A^{\prime }{u}^{\prime }+B^{\prime }{v}^{\prime }=f$. Solving gives $A^{\prime }=-vf/w$ and $B^{\prime }=uf/w$. Zero integration constants enforce both initial conditions. ThereforeFor fixed $t$, this kernel is a linear combination of homogeneous solutions, so $K_{xx}+p{K}_x+qK=0$. On the diagonal $K(t,t)=0$ and $K_x(t,t)=(v^{\prime }(t)u(t)-u^{\prime }(t)v(t))/w(t)=1$. Differentiating the integral twice now verifies the forced equation as well as $y(0)=y^{\prime }(0)=0$. Standard uniqueness of the linear initial-value problem makes this the solution, for continuous $f$.

If $u,v$ have period $T>0$, their derivatives do also, hence $w$ is periodic. The kernel is periodic in its first variable. With $f=v$, split the integral at $x$ and use this periodicity to getAt $x=0$, the right side is $-{\int }_0^{T}v(t{)}^2/w(t)dt<0$, because $w>0$ and $v^{\prime }(0)=1$ ensures $v$ is not identically zero. Thus $\boxed{y(T)\ne y(0)}$, so $y$ cannot have period $T$. This direct sign argument also rules out any apparent cancellation between the two terms.

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.

For fixed interior $t$, the Green's function vanishes at both endpoints, is continuous at $x=t$, solves the homogeneous equation away from $t$, and satisfiesThis derivative jump is the normalization required by the leading coefficient one in $L$. Finally, a homogeneous solution satisfying the first boundary condition is $C{e}^{-2x}\sin x$, and the second boundary condition at $\pi /2$ forces $C=0$. Thus the boundary value problem has exactly one solution for every continuous forcing, justifying the unique inverse represented by this kernel.

The same variation of parameters construction works on $[0,\pi ]$. All solutions satisfying the boundary condition at zero have the formAt $x=\pi$, the homogeneous term vanishes for every $C$; it can no longer be chosen to correct the final boundary value. Since $\sin (\pi -t)=\sin t$, that boundary value is zero exactly whenThis condition is necessary by the displayed general solution, and sufficient by the same formula. When it holds, the full answer isFor complex-valued forcing, allow $C\in \mathbb{C}$ instead. The freedom is addition of the nonzero homogeneous Dirichlet solution $e^{-2x}\sin x$. At this endpoint separation, the homogeneous boundary problem has a nonzero solution. The resulting Fredholm solvability condition restricts the forcing to the compatible functions described above. In the language of the Fredholm alternative, the condition is orthogonality to the adjoint null solution $e^{2x}\sin x$ of $D^2-4D+5$ with zero endpoint values.

For $\sin l\ne 0$, use variation of parameters with the homogeneous solutions $\cos x,\sin x$, whose Wronskian is one. Write $y=A(x)\cos x+B(x)\sin x$ and impose $A^{\prime }\cos x+B^{\prime }\sin x=0$. The differential equation gives $-A^{\prime }\sin x+B^{\prime }\cos x=f$, henceAfter integrating from zero and using the left boundary condition,The right boundary condition determinesso the unique solution isCombining the two integrals gives $y(x)={\int }_0^{l}G(x,t)f(t)dt$, with Green's functionIt is symmetric in $x,t$, as expected from self-adjointness, and has zero endpoint values. At $x=t$ it is continuous, while its first derivative obeys the jump conditionThus $({\partial }_x^2+1)G=\delta (x-t)$, with the positive normalization of the Dirac delta function required for $\mathcal{L}y=f$.

Use variation of parameters, writing $y_p=t_1{u}_1+t_2{u}_2$ and imposing $t_1^{\prime }{u}_1+t_2^{\prime }{u}_2=0$. Differentiation then gives $y_p^{\prime }=t_1{u}_1^{\prime }+t_2{u}_2^{\prime }$. Since the $u_j$ solve the homogeneous equation,Thus solve the two-by-two linear systemIts determinant is the nonzero Wronskian $W$. Cramer's rule gives $t_1^{\prime }=-u_{2}R/W$ and $t_2^{\prime }=u_{1}R/W$, and integration gives the stated antiderivatives. Their integration constants only add homogeneous solutions. The construction applies on a regular interval where the forcing and coefficients are defined and $W\ne 0$.

The forcing $\sec (2x)$ is undefined at $x=\pm \pi /4$, both inside the printed interval. Therefore there is no classical solution on the whole stated interval. Solve separately on the three connected components obtained by removing those two points.

Choose $u_1=\cos (2x)$ and $u_2=\sin (2x)$, whose Wronskian is two. Variation of parameters gives $t_1^{\prime }=-(1/2)\tan (2x)$ and $t_2^{\prime }=1/2$, so a particular solution is $\frac{1}{4}\cos (2x)\log |\cos (2x)|+\frac{x}{2}\sin (2x)$. The full answer on each regular component isThe constants $A,B$ can differ between $(-\pi /2,-\pi /4)$, $(-\pi /4,\pi /4)$ and $(\pi /4,\pi /2)$. Differentiation gives $y^{\prime \prime }+4y=1/\cos (2x)$ wherever the forcing is defined. Although the particular solution has finite one-sided values at the singularities, its derivative has a logarithmic divergence, and the equation itself is undefined there. It cannot be promoted to one classical solution across the poles.

Two functions are linearly dependent if constants $c_1,c_2$, not both zero, satisfy $c_1{y}_1+c_2{y}_2=0$ identically. Differentiating also gives $c_1{\dot{y}}_1+c_2{\dot{y}}_2=0$. Since the determinant of these two equations is the nonzero Wronskian $W$, both constants must vanish. Hence the two solutions are linearly independent.

Differentiating the Wronskian and substituting the homogeneous linear differential equation givesIntegrating proves Abel's identity:For variation of parameters, impose ${\dot{a}}_1{y}_1+{\dot{a}}_2{y}_2=0$ on the proposed particular solution. Then $\dot{y}=a_1{\dot{y}}_1+a_2{\dot{y}}_2$, and substitution into the inhomogeneous differential equation leaves ${\dot{a}}_1{\dot{y}}_1+{\dot{a}}_2{\dot{y}}_2=f$. Solving this two-by-two linear system givesArbitrary constants of integration merely add a homogeneous solution.

For the constant positive coefficient $q={\omega }^2$, take $\omega >0$. Direct differentiation verifies that $y_1=\cos \omega t$ and $y_2=\sin \omega t$ solve the homogeneous linear differential equation; their Wronskian is $W=\omega$. The requested integrands areTaking $t_0=0$ yieldsThus $|a_1(t)|\sim t/(2\omega )$ grows with power one, whereas $a_2$ is bounded. Their combination simplifies to $y_p=-t\cos (\omega t)/(2\omega )+\sin (\omega t)/(2{\omega }^2)$; the final sine term is itself a homogeneous solution. The secular term is the response to resonance in a differential equation.

The Wronskian satisfies Abel identity. Differentiating and substituting the homogeneous equations givesso $W(x)=-\exp [-{\int }_0^{x}p(r)dr]\ne 0$. The two homogeneous solutions are therefore independent.

Define the variation of parameters kernelFor fixed $s$, it satisfies the homogeneous differential equation in $x$. Direct evaluation gives $K(s,s)=0$ and $K_x(s,s)=1$. For $y(x)={\int }_0^{x}K(x,s)f(s)ds$, differentiation under the integral therefore yieldsCombining these identities proves $y^{\prime \prime }+p{y}^{\prime }+qy=f$; both initial values vanish. Continuity of the coefficients gives uniqueness of this initial value problem, establishing the proposed formula rather than merely recognizing it as a standard method.

For the oscillator, $y_1=\sin x$, $y_2=\cos x$ and $W=-1$, so the kernel is $\sin (x-s)$. HenceUsing ${\int }_0^x\cos s\sin sds={\sin }^{2}x/2$ and ${\int }_0^x{\sin }^{2}sds=x/2-\sin (2x)/4$ givesThe factor $x\cos x$ reflects resonance with the homogeneous oscillator frequency.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.

Write $W(s)=a_1\mathrm{Ai}(Z(s))+a_2\mathrm{Bi}(Z(s))$, using the independent Airy functions. To solve $(D^2-1)\hat{v}=W$, the variation of parameters formula, equivalently a direct convolution, gives the velocity reconstruction from the Orr-Sommerfeld vorticity:The reference point $y_0$ is arbitrary. Twice differentiating the integral produces $W(y)$ because $\sinh 0=0$ and its derivative at zero is one, verifying the formula. The four constants are independent: applying $D^2-1$ to an identically zero combination first forces $a_1=a_2=0$, then the complementary solutions force $C_{+}=C_{-}=0$. The boundary conditions would be $\hat{v}(y_j)=D\hat{v}(y_j)=0$, $j=1,2$. Their compatibility selects the eigenvalues; they need not be imposed here.

Resolve the forcing in the eigenbasis:In the zero-eigenvalue direction, a constant forcing produces the particular integral $(3a-1)t{u}_1$. In the unit-eigenvalue direction, a constant particular integral is $-(1-2a)u_2$. Thus a particular integral depends on time exactly whenAdding the complementary solution from part (a), the general solution isThis is the eigenvector form of the usual variation of parameters calculation.

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.

For a radial function in three dimensions, $(\Delta -1)y=g(x)$ becomes $y^{\prime \prime }+2y^{\prime }/x-y=g(x)$. Setting $u=xy$ gives $u^{\prime \prime }-u=xg(x)$, reducing it to a constant-coefficient linear ordinary differential equation. If $g=e^{-3x}/x^2$, a decaying particular solution obtained by variation of parameters isThe small-argument expansion of the exponential integral givesThe undetermined decaying homogeneous coefficient $\alpha$ is fixed by a boundary condition or by a matched asymptotic expansion.

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.