Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 3

The substitution $y=e^{x}g$ reduces the homogeneous linear differential equation toA first integration gives $g^{\prime }=Cx{e}^{-x}$, and a second gives $g=D-C(x+1)e^{-x}$. HenceThe Wronskian of $1+x,e^x$ is $x{e}^x$, so they are linearly independent on either side of zero. At zero the differential equation has a regular singular point. The displayed solutions extend analytically through it; their linear combinations describe the global solutions when two continuous derivatives are required there. If no matching condition is imposed, the two sides have independent constants.

Use the boundary-adapted homogeneous solutionsFor a source point $\xi \ne 0$, the Green function isIt vanishes at both endpoints and is continuous at $x=\xi$. Its first derivative has the jumpas required for a monic second-order differential operator with a Dirac delta source. Away from that source it is a homogeneous solution and extends smoothly through $x=0$.

The interior singular point requires a qualification. This is a Green kernel with a vanishing Wronskian at an interior singular point, with a $1/\xi$ singularity in its source variable. It is not an ordinary integral inverse for unrestricted forcing. For smooth forcing $f(\xi )=\xi q(\xi )$, however, the product $Gf$ extends continuously at zero and the integral is well defined. This includes part (iii). We use solutions with two continuous derivatives across zero, interpreting the equation there after multiplication by $x$. For such a solution, expansion at zero forces $f(0)=0$; arbitrary nonzero forcing at zero cannot satisfy this regularity. Merely imposing one continuous derivative would leave a piecewise homogeneous ambiguity, so the original singular coefficients alone do not specify an unrestricted boundary-value inverse.

The source factor cancels the singular denominator of the Green function. Splitting the integral at $\xi =x$ givesUsing $\int (1+\xi )e^{-\xi }d\xi =-(\xi +2)e^{-\xi }$, these two integrals are $e-(x+2)e^{-x}$ and $5/2-x-e(x+2)e^{-x}/2$, respectively. Simplification givesThe exponential and linear terms are homogeneous solutions, while direct differentiation gives $L(-x^2)=x$ for $x\ne 0$. Also $y(-1)=y(1)=0$, and the expression is analytic through zero. Uniqueness in this regular class follows because any difference is a global linear combination $A{e}^x+B(1+x)$: the left boundary forces $A=0$, and the right then forces $B=0$.