= Solution
By definition, the <relaxation modulus> is the <stress> generated by a unit step of <strain> from an initially relaxed state. Set $\epsilon(t)=H(t)$, where $H$ is the <Heaviside step function>, and write $\sigma(t)=G(t)$. Integrating the <linear ordinary differential equation> across the step shows that neither $G$ nor $G'$ can jump: a jump would produce a <Dirac delta function> or its <derivative>, absent from the forcing. Hence $G(0^+)=G'(0^+)=0$.
For $t>0$, a particular solution is $1$. The <characteristic equation> factors as
$$
15\lambda^2+8\lambda+1=(5\lambda+1)(3\lambda+1)=0,
$$
so $G=1+A e^{-t/5}+B e^{-t/3}$. The initial conditions give $A+B=-1$ and $A/5+B/3=0$, whence
$$
\boxed{G(t)=\left(1-\frac52e^{-t/5}+\frac32e^{-t/3}\right)H(t).}
$$
The time and modulus units are those implicit in the printed coefficients. A <Laplace transform> check gives $\widehat G(s)=1/[s(5s+1)(3s+1)]$.
The original PDF really does put undifferentiated <strain> on the right-hand side. The resulting response rises from zero to one, with $G'(t)=\tfrac12(e^{-t/5}-e^{-t/3})>0$, rather than exhibiting ordinary <stress relaxation>. There is a stronger physical problem if this is intended as a passive material law: with the convention $e^{i\omega t}$ the <stress>-to-<strain> amplitude ratio is
$$
G^*(\omega)=\frac1{1-15\omega^2+8i\omega},\qquad
G''(\omega)=\frac{-8\omega}{(1-15\omega^2)^2+64\omega^2}<0\quad(\omega>0).
$$
The <loss modulus> would give negative cycle-averaged supplied power $\langle\sigma\dot\epsilon\rangle=\omega G''|\epsilon_0|^2/2$, contrary to <passivity in linear viscoelasticity>. The boxed expression is therefore the formal <relaxation modulus> of the equation as printed, not a claim that this equation describes an ordinary passive <viscoelasticity> model. No derivative of the input has been silently inserted.
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