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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 $y=\alpha x+\beta x^2$, 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.

A sufficient condition for existence of the Laplace transform is that $y$ be piecewise continuous on each bounded interval and have exponential order: $|y(t)|\le C{e}^{at}$ for sufficiently large $t$. The defining integral then converges absolutely for real $p>a$, and more generally defines a holomorphic function for complex $p$ with $\mathrm{Re}p>a$. The boundary $p=a$ may or may not be included, depending on $y$.

Write $Y(p)=\mathcal{L}[y](p)$. To solve the differential equation by Laplace transform, initially seek a sufficiently regular solution with $y(0)=1$ and finite $y^{\prime }(0)=b$. The Laplace transform of a derivative givesUsing $\mathcal{L}[ty]=-Y^{\prime }$ from part (i), the transformed differential equation becomesorAfter division by $p(p+1)$, an integrating factor is $p^2(p+1)$, since $(3p+2)/(p(p+1))=2/p+1/(p+1)$. ConsequentlyFor a bounded $y$, $|Y(p)|\le \Vert y{\Vert }_{\infty }/p$ for $p>0$. Letting $p\downarrow 0$ forces $C=0$, an instance of how boundedness selects a Laplace transform at a singular endpoint. By the exponential formula in part (iii), this yieldsDirect substitution checks the differential equation, boundedness and the required limit.

To complete the uniqueness argument, no unproved endpoint regularity is needed. The two functions $e^{-t}$ and $t-1$ solve the homogeneous linear differential equation on $(0,\infty )$, and their Wronskian is $t{e}^{-t}\ne 0$. They form a fundamental system of solutions on that interval, so every classical solution is $A{e}^{-t}+B(t-1)$. Boundedness forces $B=0$ and the limit at zero forces $A=1$. Thus the bounded solution with the specified limit is unique. The limit alone would not suffice: $e^{-t}+C(t-1+e^{-t})$ has the same limit for every $C$, but is unbounded when $C\ne 0$.

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.

In the Laplace integral method for a differential equation, differentiate the integral three times with respect to $z$, and replace $z{e}^{z\zeta }$ by ${\partial }_{\zeta }{e}^{z\zeta }$. Integration by parts givesThus the weight equation for this construction and its boundary condition arewhich giveThese are the conditions obtained by eliminating the transformed interior term and the endpoint term separately. As usual for this method, differentiation and integration by parts need convergence assumptions; the contours constructed next make every such step valid locally uniformly for all complex $z$. A literal necessity claim for arbitrary weights and contours would additionally require uniqueness of the chosen representation by a Laplace integral; a null weight in a contour Laplace representation can otherwise be added without changing the solution.

LetThe quartic decay sectors for a contour integral contain the four rays $\zeta =r{u}_j$. Along each ray, the quartic term is ${\zeta }^4/4=-r^4/4$. For $j=1,2,3$, choose ${\Gamma }_j$ to travel from infinity on ray $u_0$ to zero, and then from zero to infinity on ray $u_j$. Set $A=1$ and defineOn any compact set of $z$ values, the integrand and each of its $z$ derivatives are bounded by a constant times $r^m{e}^{-r^4/4+Mr}$ for some $M$. These bounds are integrable at infinity, so differentiation under the integral sign is justified and the functions are entire. The endpoint term vanishes at both infinite ends. There is no extra term at zero because the two pieces join with the same integrand value and opposite endpoint orientations. Therefore all three functions solve the differential equation.

To prove independence, evaluate their first three initial data. For $m=0,1,2$,If ${\sum }_{j=1}^3{c}_j{w}_j=0$, these initial data give ${\sum }_{j=1}^3{c}_j(u_j^{\ell }-u_0^{\ell })=0$ for $\ell =1,2,3$. Define $c_0=-{\sum }_{j=1}^3{c}_j$. ThenThe coefficient matrix is a Vandermonde matrix at four distinct fourth roots of $-1$, and hence is invertible. All $c_j$ vanish. There is also a concrete reason for the earlier qualification about necessity. Deform the incoming ray of ${\Gamma }_1$ to angle $\pi /3$ and its outgoing ray to angle $2\pi /3$, within their respective quartic decay sectors. The quartic integral is unchanged and remains nonzero. On this deformed contour, the extra entire weight $e^{{\zeta }^2}$ contributes zero for every $z$: throughout the closing wedge, $\mathrm{Re}({\zeta }^2)\le -r^2/2$, so the large closing arc tends to zero and the Cauchy integral theorem gives a zero integral. Thus $g=e^{{\zeta }^4/4}+e^{{\zeta }^2}$ represents the same entire solution and has vanishing endpoint terms, although it does not satisfy $g^{\prime }={\zeta }^{3}g$. The canonical weight equation is the standard sufficient construction, rather than an unconditional uniqueness assertion about contour representations.

Thus the three contours give linearly independent nonzero solutions, and they form a fundamental system of solutions for this third-order linear equation. This proves independence directly instead of inferring it merely from the presence of three different contours.

The characteristic polynomial is $r^2+2r+2=(r+1{)}^2+1$, with distinct complex roots $r=-1\pm i$. Taking the real and imaginary parts of the corresponding exponential functions gives the real fundamental system of solutions $e^{-x}\cos x$ and $e^{-x}\sin x$. Consequently the general real solution isTheir Wronskian is $e^{-2x}\ne 0$, so they span the two-dimensional real solution space.