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In the classical risk model, let $N(t)$ be the Poisson process of claim arrivals, independent of the independent claim sizes. The reserve starts at $u$ and earns premiums continuously at rate $c$, soHere $\psi$ is the ultimate ruin probability and $\varphi =1-\psi$ is the survival probability. The relative safety loading is $\theta =(c-\lambda \mu )/(\lambda \mu )>0$.

To derive the survival integro-differential equation, condition on the first claim time $T$, which has the exponential distribution of rate $\lambda$. Immediately before that claim the available capital is $u+cT$. A claim $x\le u+cT$ leaves survival probability $\varphi (u+cT-x)$, since the future claim process restarts with the same independent law; a larger claim causes ruin immediately. The first-claim decomposition for survival probability is thereforePut $r=\lambda /c$ and $H(v)={\int }_0^v\varphi (v-x)f(x)dx$. Changing variables to $v=u+ct$ yieldsThis representation justifies differentiation. The function $H$ is continuous: extend $\varphi$ by zero to negative arguments, then use continuity of translations of the integrable density $f$ in $L^1$ to control its convolution with the bounded function $\varphi$. Differentiating the lower endpoint of the displayed integral gives ${\varphi }^{\prime }=r\varphi -rH$. Since $c=(1+\theta )\lambda \mu$,This deduction uses the first-arrival law rather than taking the desired differential equation as an assumption.

For $f(x)=4x{e}^{-2x}$, the claim law is an Erlang distribution of shape two and rate two. Its mean is $\mu =4{\int }_0^{\infty }{x}^2{e}^{-2x}dx=1$. With $\theta =2$, we have $r=1/3$. Changing the convolution variable to $t=u-x$ givesHence the first requested specialization isLet $J(u)={\int }_0^u\varphi (t)e^{2t}dt$. Differentiation under the finite integral gives $I^{\prime }=J$ and $J^{\prime }=\varphi e^{2u}$. Differentiating the preceding equation and using $4e^{-2u}I=\varphi -3{\varphi }^{\prime }$ then givesThus the second identity isDifferentiate once more:The preceding second-derivative equation gives $8e^{-2u}J=-6{\varphi }^{\prime \prime }-10{\varphi }^{\prime }+4\varphi$. Substitution cancels the undifferentiated terms and provesThis is the Erlang claim-size differential equation for survival probability with $\beta =2$ and $r=1/3$.

Its characteristic polynomial factors as $z(3z+8)(z+1)$. The limit $\varphi (u)\to 1$ therefore makes the general bounded solutionThe supplied zero-capital survival probability is $\varphi (0)=2/3$, so $A+B=-1/3$. We also need the initial derivative carried by the original integral equation: $I(0)=0$ implies ${\varphi }^{\prime }(0)=\varphi (0)/3=2/9$, giving $-A-(8/3)B=2/9$. Solving gives $A=-2/5$ and $B=1/15$. Therefore the survival and ruin probabilities areThe extra initial derivative is essential: $\varphi (0)$ and the limit at infinity alone would leave one free parameter in the third-order equation. The result has $\psi (0)=1/3$ and tends to zero. It is positive and decreasing, since ${\psi }^{\prime }(u)=-(2/5)e^{-u}+(8/45)e^{-8u/3}<0$ for $u\ge 0$. These signs agree with its interpretation as an ultimate ruin probability.