Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 2 / 1B / Solution

For a straight-line solution $y=mx$, substitution into the differential equation gives $m=(1-2m)/(2+m)$. Thus $m^2+4m-1=0$, and the two possible slopes areThese straight lines solve the differential equation on either side of the origin. At the origin itself the right-hand side is undefined.

An isocline of slope $k$ has equation $(1-2k)x-(2+k)y=0$. For $k\ne -2$ it is the straight line $y=(1-2k)x/(2+k)$; for $k=-2$ it is $x=0$, excluding the origin. In particular, $y=x/2$ is the zero-slope isocline. The line $y=-2x$ is excluded from the domain of the differential equation; slopes tend to infinity on approaching it. Draw the flow vectors proportional to $(1,(x-2y)/(2x+y))$, with positive horizontal component. This specifies orientation by increasing $x$, rather than by an auxiliary time variable.

To integrate, put $y=xu$ where $x\ne 0$. ThenThe left-hand integral is $-\frac{1}{2}\log |1-4u-u^2|$, so the nonconstant-$u$ solutions have the first integralThe constant-$u$ solutions omitted by separation are exactly the two straight lines already found, corresponding to $C=0$. Alternatively, differentiating this first integral gives $(2y+4x)y^{\prime }+4y-2x=0$, which recovers the original differential equation wherever $2x+y\ne 0$ and also verifies continuation through $x=0$ when $y\ne 0$.

The first integral describes hyperbolas with the two straight-line solutions as asymptotes. For $C=1$, the upper branch crosses the vertical axis at $(0,1)$ and exists for every real $x$. For $C=-1$, each branch is confined to $|x|\ge 1/\sqrt{5}$ and reaches a vertical tangent on the excluded line $y=-2x$; the corresponding classical solution intervals exclude those endpoints. These give the two distinct types of curves requested in the sketch.

The initial condition selects $C=1$ and the upper branch, henceHere $2x+y=\sqrt{5x^2+1}>0$, so this really is a global solution of the initial value problem, and $y^{\prime }(0)=-2$ agrees with the right-hand side.