= Solution
Use the <classical risk model> surplus $U_t=u+ct-\sum_{j=1}^{N_t}X_j$, with $c>0$, and write $\varphi$ for its ultimate <survival probability>. The <first-claim decomposition for survival probability> follows by conditioning on the first arrival time of the <Poisson process>. That time has <exponential distribution> of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\leq u+ct$, after which the <Markov property> restarts the same risk model. Consequently
$$
\varphi(u)=\int_0^\infty\lambda e^{-\lambda t}
\int_0^{u+ct}\varphi(u+ct-x)f(x)\,dx\,dt.
$$
Put $C(v)=\int_0^v\varphi(v-x)f(x)\,dx$ and change variable $v=u+ct$:
$$
\varphi(u)=\frac\lambda c e^{\lambda u/c}
\int_u^\infty e^{-\lambda v/c}C(v)\,dv.
$$
The <convolution> $C$ is continuous since $\varphi$ is bounded and $f$ is integrable. Differentiating proves the <survival integro-differential equation>
$$
\boxed{\varphi'(u)=\frac\lambda c\varphi(u)-\frac\lambda c
\int_0^u\varphi(u-x)f(x)\,dx,\qquad a_1=\lambda/c,\quad a_2=-\lambda/c.}
$$
For the specified claim law, expand its <probability density function> as $f(x)=\tfrac12e^{-x}+\tfrac14e^{-x/2}$. This is an equal-weight <mixture distribution> of <exponential distributions> with rates $1$ and $1/2$. In the <convolution>, set $z=u-x$ to obtain
$$
C(u)=\frac12e^{-u}I_1(u)+\frac14e^{-u/2}I_2(u).
$$
Since $\lambda/c=10/21$, the requested coefficients are
$$
\boxed{b_1=\frac{10}{21},\qquad b_2=-\frac5{21},\qquad b_3=-\frac5{42}.}
$$
For the <exponential-mixture differential equation for survival probability>, define $A(u)=e^{-u}I_1(u)$, $B(u)=e^{-u/2}I_2(u)$ and $r=\lambda/c$. Differentiating the integrals yields
$$
A'=\varphi-A,\qquad B'=\varphi-\tfrac12B,\qquad
\varphi'=r\varphi-\tfrac r2A-\tfrac r4B.
$$
Apply $(D+1)(D+\tfrac12)$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, giving
$$
\varphi'''+\frac32\varphi''+\frac12\varphi'
=r\varphi''+\frac{3r}{4}\varphi'.
$$
Thus
$$
\boxed{\varphi'''+\frac{43}{42}\varphi''+\frac17\varphi'=0,
\qquad c_1=43/42,\quad c_2=1/7.}
$$
The <characteristic polynomial> factors as
$$
D\left(D^2+\frac{43}{42}D+\frac17\right)
=D(D+\tfrac16)(D+\tfrac67).
$$
The <ordinary differential equation> therefore has solution $C_0+C_1e^{-u/6}+C_2e^{-6u/7}$.
The claim <expected value> is $\tfrac12\cdot1+\tfrac12\cdot2=3/2$. The <relative safety loading> is $\rho=2.1/(3/2)-1=2/5$, and the given <zero-capital survival probability> is $\varphi(0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original <survival integro-differential equation>: its <convolution> vanishes at zero, so $\varphi'(0)=(10/21)(2/7)=20/147$. Therefore
$$
C_1+C_2=-\frac57,\qquad
-\frac{C_1}{6}-\frac{6C_2}{7}=\frac{20}{147},
$$
which gives $C_1=-20/29$ and $C_2=-5/203$. The final <survival probability> and <ultimate ruin probability> are
$$
\boxed{\varphi(u)=1-\frac{20}{29}e^{-u/6}-\frac5{203}e^{-6u/7},}
$$
$$
\boxed{\psi(u)=\frac{20}{29}e^{-u/6}+\frac5{203}e^{-6u/7}\qquad(u\geq0).}
$$
Both exponential coefficients of $\psi$ are positive. Hence $\varphi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.
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