Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 14C / Solution

The chain rule gives $z_x=\sin 2x$, $z_x^2=4z(1-z)$, and $z_{xx}=2(1-2z)$. Thereforewhich, after division, is the printed equation. It has only three regular singular points, $0,1,\infty$, and is thus a Papperitz equation. At zero and one the indicial roots are $0,1/2$. At infinity, inserting $w\sim z^{-\rho }$ gives $\rho =\pm n/2$. Its Papperitz symbol isThe exponent sum is one, as required for a three-singularity second-order equation.

Multiplying the equation by $z(1-z)$ puts it in Gauss hypergeometric equation form with $a=n/2$, $b=-n/2$, $c=1/2$. The solution analytic at zero and normalized to one is ${}_2{F}_1(a,b;c;z)$. Composing it with $z={\sin }^{2}x$ gives an even analytic solution of $w_{xx}+n^{2}w=0$ with $w(0)=1$, $w_x(0)=0$. Uniqueness of this ordinary differential equation impliesThis Gauss hypergeometric function identity initially holds near zero, and on $|x|<\pi /2$ with the usual real branch; it extends by matching analytic continuation. It is not an identity on every real $x$ if the left side is always reevaluated on its principal $z$ branch: for $n=1$ that principal value is $\sqrt{1-{\sin }^{2}x}=|\cos x|$. Continuing the branch consistently restores the stated cosine.