= 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 are
$$
\boxed{m_\pm=-2\pm\sqrt5.}
$$
These 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 $x\ne0$. Then
$$
xu'=\frac{1-4u-u^2}{2+u},\qquad
\int\frac{2+u}{1-4u-u^2}\,du=\int\frac{dx}{x}.
$$
The left-hand <integral> is $-\tfrac12\log|1-4u-u^2|$, so the nonconstant-$u$ solutions have the <first integral>
$$
\boxed{y^2+4xy-x^2=C.}
$$
The constant-$u$ solutions omitted by separation are exactly the two straight lines already found, corresponding to $C=0$. Alternatively, differentiating this <first integral> gives $(2y+4x)y'+4y-2x=0$, which recovers the original <differential equation> wherever $2x+y\ne0$ and also verifies continuation through $x=0$ when $y\ne0$.
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 $|x|\geq1/\sqrt5$ 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, hence
$$
\boxed{y(x)=-2x+\sqrt{5x^2+1},\qquad -\infty<x<\infty.}
$$
Here $2x+y=\sqrt{5x^2+1}>0$, so this really is a global solution of the <initial value problem>, and $y'(0)=-2$ agrees with the right-hand side.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-2-isoclines.png]
{title=Isoclines, increasing-x flow vectors and the two types of solution hyperbolas}
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