Past exam of the mathematical methods I course of the University of Oxford / 2021 / 10

Work over the reals with $x\ne 0$, and put $p=dy/dx$. The algebraic equation implies $p\ne 0$ andDifferentiate this identity and use $y^{\prime }=p$; cancellation gives $dp/dx=p/x$. Hence $p=Cx$, with nonzero real $C$, and substitution gives the complete familyThere is no extra real singular branch lost in the cancellation, since $p^2+1\ne 0$. At a point $(x_0,y_0)$ with $x_0\ne 0$, the permitted constants satisfy $x_0^2{C}^2-2y_{0}C-1=0$, givingThese are distinct, real and nonzero, so exactly two integral curves pass through each such point. This does not contradict an Picard-Lindelöf theorem: the implicit equation permits two different slopes and is not one specified explicit first-order ordinary differential equation. Each choice of the smooth slope branch separately defines a locally unique curve.

The two real slope roots of $p^2+fp-1=0$ are distinct because their discriminant is $f^2+4>0$. Vieta's formulas give $p_{+}{p}_{-}=-1$, so the tangent vectors $(1,p_{+})$ and $(1,p_{-})$ have zero scalar product. Whenever the coefficients define local solution curves, the two families are therefore orthogonal trajectories.

For the stated example at $(3,4)$, the slopes are $3$ and $-1/3$. In the family from part a, the constants are $1$ and $-1/9$, givingBoth curves pass through $(3,4)$, and $3(-1/3)=-1$ verifies the right angle.