Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 4

Assume the homogeneous function is differentiable. Applying the multivariable chain rule to $f(tx,ty)=t^{m}f(x,y)$ gives, for $t$ near $1$,Setting $t=1$ proves Euler theorem for homogeneous functions:Only positive $t$ near $1$ are needed, so the derivation remains valid even when a noninteger $m$ makes the printed phrase “all real $t$” require qualification.

The linear ordinary differential equation has differential operator $(D-3{)}^2$. Put $y=e^{3x}v$ and use the product rule twice:Substitution cancels the $v$ and $v^{\prime }$ terms, leaving $v^{\prime \prime }=4$. Two integrations give $v=2x^2+C_{1}x+C_2$. Hence the general solution isThe two arbitrary constants account for the order of the differential equation; the $x^2$ factor in the particular term reflects the repeated root $3$ of the characteristic polynomial.

On an interval with $x\ne 0$, set $y=xv$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution givesThis is a separable differential equation. Since $1+v^2>0$ for real $v$, separation loses no real constant solution:Thus the general solution on any interval avoiding zero and the poles of the tangent function isAt $x=0$ the original differential equation forces $y(0)=0$, but no real solution from this family extends over a whole interval through zero: $\log |x|+C$ encounters infinitely many poles as $x\to 0$.

For $x\ne \pm 1$, write the linear ordinary differential equation asAn integrating factor on each interval avoiding $\pm 1$ is $\mu (x)=1/(1-x^2)$; multiplication by a nonzero constant, including a sign, is immaterial. ThereforeThe partial fraction decomposition isIntegrating gives the general solutionThe constant is independent on each connected interval of definition. At the singular points the undivided differential equation forces $y(1)=0$ and $y(-1)=1$. The displayed family tends to these values, but its derivative has a logarithmic divergence there, so no member extends as a differentiable function through either singular point.