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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.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.