Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 14 / c

Let $F(p)=2p/(1-p^2)$, so that $y=xF(p)$. Differentiation givesMove the first term to the left and useSince $1+p^2>0$ for real $p$ and $p\ne \pm 1$, cancellation givesThis is the slope parameter for a homogeneous first-order equation reduction.

For $p\ne 0,\pm 1$, separate variables and use partial fraction decomposition:Integration yieldsOn any connected solution interval, absorb the fixed signs into a real constant to writeThe two real signs of the square root are chosen consistently on a branch where $C/(x+C)>0$. The constant solutions of the reduced equation are $p=0,1,-1$. The original equation excludes the last two; $p=0$ is retained by $C=0$ for $x\ne 0$.

From $p^2=C/(x+C)$, we have $1-p^2=x/(x+C)$. Substitution into the original relation gives $y=2(x+C)p$, soThese are real parabola branches, with either sign of $C$ allowed when the branch is real. For $C\ne 0$, differentiating gives $p=2C/y$ and verifies both the relation and the slope. Use graph intervals with $y\ne 0$ and $x\ne 0$, so the slope is finite and $p\ne \pm 1$. The separate constant-slope case $p=0$ gives $y=0$, also represented by the degenerate choice $C=0$. Squaring has not imposed a sign on $y$, so both real graph branches must be included.