 Solution

By definition, the relaxation modulus is the stress generated by a unit step of strain from an initially relaxed state. Set $ϵ(t)=H(t)$, 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 $G^{\prime }$ can jump: a jump would produce a Dirac delta function or its derivative, absent from the forcing. Hence $G(0^{+})=G^{\prime }(0^{+})=0$.

For $t>0$, a particular solution is $1$. The characteristic equation factors asso $G=1+A{e}^{-t/5}+B{e}^{-t/3}$. The initial conditions give $A+B=-1$ and $A/5+B/3=0$, whenceThe time and modulus units are those implicit in the printed coefficients. A Laplace transform check gives $\hat{G}(s)=1/[s(5s+1)(3s+1)]$.

The original PDF really does put undifferentiated strain on the right-hand side. The resulting response rises from zero to one, with $G^{\prime }(t)=\frac{1}{2}(e^{-t/5}-e^{-t/3})>0$, 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 isThe loss modulus would give negative cycle-averaged supplied power $⟨\sigma \dot{ϵ}⟩=\omega G^{\prime \prime }|{ϵ}_0{|}^2/2$, 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.