Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 13

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

For this first-order linear differential equation, an integrating factor on an interval avoiding zeros of $\cos x$ is $\cos x$. The product rule then givesThe initial condition gives $3/\sqrt{2}=1/\sqrt{2}+C$, hence $C=\sqrt{2}$. ThusThe displayed interval is the maximal one containing the specified initial point, since the coefficient $\tan x$ is undefined at its endpoints.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

The initial condition gives $C=\sin (\pi /3)=\sqrt{3}/2$. Since $\pi /3$ belongs to the principal inverse sine range, choose $k=0$ in part (i). The particular solution isThe upper endpoint is excluded because $y/x\to \pi /2$ there and the original tangent ceases to be defined.

Again $C=\sin (2\pi /3)=\sqrt{3}/2$, but the principal inverse sine would give $\pi /3$ at $x=1$. The required branch is $v=\pi -\arcsin (Cx)$, corresponding to $k=1$. ThereforeAt $x=1$ the bracket is $2\pi /3$, as required. Both the initial condition and the original differential equation are thus satisfied on the stated branch.