Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 15

On a nonconstant branch with $\sin y\ne 0$, regard $x$ as a function of $y$. The equation becomes the linear ordinary differential equationAn integrating factor is $\sin y$, givingTaking an antiderivative yields the implicit familyOn a regular branch this can also be written $x=\sin y+C/\sin y$. The latter form loses the constant solutions $y=n\pi$, which must be restored separately.

The implicit relation is a first integral of an ordinary differential equation: along any differentiable solution of the original equation,Conversely, a differentiable graph satisfying this relation solves the original equation wherever $\cos y\ne 0$. The original equation is undefined when $\cos y=0$; points at which $2\sin y-x=0$ also need separate attention before solving explicitly for $y^{\prime }$. This is why dividing immediately by $\tan y$ can lose valid solutions and obscure nonuniqueness.

The initial data force $C=0$, so the implicit relation factors as $\sin y(x-\sin y)=0$. Near $(0,0)$ this permits either the constant branch or the sine branch:Both satisfy the original equation, including its degenerate value $0=0$ at the initial point. They have different slopes there, zero and one, so the initial value problem is not unique. There are no further differentiable local branches through that point: away from $x=0$ the two factors cannot switch, and joining the two different branches at zero would give unequal one-sided derivatives.

The nonconstant branch is defined as a solution for $-1<x<1$; the constant branch exists for every real $x$.

Again $C=0$. Since the initial sine is nonzero, the local branch must satisfy $\sin y=x$. Selecting the inverse-sine branch through $y=\pi /6$ givesThis passes through the specified point and has $y^{\prime }=1/\sqrt{1-x^2}$, which verifies the original equation on this interval.

Here $C=0$ as well, but the branch through $5\pi /6$ is the other inverse-sine branch:Its derivative is $-1/\sqrt{1-x^2}$, and both the sign of $\tan y$ and the original equation agree with this choice.

The characteristic equation of the associated homogeneous linear differential equation is $m^2+7m+12=(m+3)(m+4)=0$. A constant particular integral is $1/2$, soThe first boundary condition gives $A+B=-1/2$. The second gives $A{e}^{-1}+B{e}^{-4/3}=-e^{-1}/2$. Eliminating $A$ yields $B(e^{-4/3}-e^{-1})=0$, hence $B=0$ and $A=-1/2$. Therefore