Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 1

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.

The product rule gives $y^{\prime }=x^n{z}^{\prime }+n{x}^{n-1}z$ and $y^{\prime \prime }=x^n{z}^{\prime \prime }+2n{x}^{n-1}{z}^{\prime }+n(n-1)x^{n-2}z$. On an interval avoiding $x=0$, substitution and division by $x^{n+2}$ giveFor a constant-coefficient differential equation, the coefficient of $z^{\prime }$ forces $2n+4=0$. At $n=-2$, the remaining coefficient $n^2+3n+2$ also vanishes. ThusThe characteristic equation of a constant-coefficient differential equation is $r^2+2r+2=0$, with complex conjugate roots of a polynomial $-1\pm i$. Its real solutions $e^{-x}\cos x$ and $e^{-x}\sin x$ are linearly independent. Therefore the general solution isThe constants can be chosen independently on the two sides of the singular point. The power substitution reducing a differential equation to constant coefficients here is invertible wherever $x\ne 0$.

Use the change of variables $u=x+y$. Then $u^{\prime }=1+y^{\prime }=1+\cos u=2{\cos }^2(u/2)$, a separable differential equation. Where $u$ is not an odd multiple of $\pi$, separation of variables givesIndeed the derivative of $\tan (u/2)$ with respect to $u$ is $1/(1+\cos u)$. The nonconstant branches can also be written $y=2\arctan (x+C)+2k\pi -x$, $k\in \mathbb{Z}$. Dividing by $1+\cos u$ excluded its zeros: the missing equilibrium solutions for $u$ give the additional solutions of a differential equationThey satisfy $y^{\prime }=-1=\cos (x+y)$ directly. The transformed right side is continuously differentiable, so the uniqueness theorem for ordinary differential equations prevents a nonconstant branch from crossing an equilibrium solution at finite $x$.