Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 7

With $y=xv$, direct differentiation givesThe homogeneous ordinary differential equation therefore has general solution $y=x(A\cos x+B\sin x)$. Chooseso $y_1(\pi /2)=0$ and $y_2(\pi )=0$. Their Wronskian in the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ is $W(x)=x^2$. There is no nonzero homogeneous solution satisfying both boundary conditions, since the first forces $B=0$ and the second forces $A=0$.

For the operator as printed, whose leading coefficient is $x^2$, the Green's function isIt satisfies the endpoint boundary conditions, is continuous across $x=s$, and its first derivative has jump $1/s^2$. Multiplying that jump by the leading coefficient $s^2$ produces the unit source required by a Green's function. This factor matters because the operator was not normalized to leading coefficient one; it is an instance of a Green function under dependent-variable rescaling.

Thus $y(x)={\int }_{\pi /2}^{\pi }G(x,s)f(s)ds$. For $f(s)=s^3$, split the integral at $s=x$ to obtainThereforeThe substitution above gives $v^{\prime \prime }+v=1$, so the forcing is indeed $x^3$, and both endpoint values vanish.