I was given a homework question
Suppose a bone described by the Kelvin model is deformed by a force
$$F(t)=F_0(\Theta(t)-\Theta(t-\tau)+\Theta(t-2\tau)),$$
where $\Theta$ is the Heaviside function and $F_0 \in \mathbb{R}$ is a constant. Find an analytical expression for the response deformation $x(t)$ and sketch it.
Below is what I have done so far, but I'm confused on how to proceed with solving the ODE for the more complicated terms as well as how to combine solutions, especially as we have not looked at using Laplace transforms yet.
Suppose a bone described by the Kelvin model is deformed by a force $$F(t)=F_0\left(\Theta(t)-\Theta(t-\tau)+\Theta(t-2\tau)\right), $$ where $\Theta$ is the Heaviside function. Then the response deformation $x$ satisfies the differential equation $$\displaystyle F+\tau_{\epsilon}\frac{dF}{dt}=K_2\left(x+\tau_{\sigma}\frac{dx}{dt}\right). $$
Rewriting using the equation for $F$ gives $$\begin{align} \displaystyle x'+ \frac{1}{\tau_{\sigma}}x & = \displaystyle \frac{F_0}{K_2\tau{\epsilon}}\left(\Theta(t)-\Theta(t-\tau)+\Theta(t-2\tau)\right) \\ & + \displaystyle \frac{F_0\tau_\epsilon}{K_2\tau_\sigma}\left(\delta(t)-\delta(t-\tau)+\delta(t-2\tau)\right) \end{align} $$
For $t\leq 0$, the equation becomes $$\displaystyle x'+ \frac{1}{\tau_{\sigma}}x=0 $$ which has solution $$x(t)=\displaystyle C_1\exp\left({\frac{t}{\tau_\sigma}}\right).$$
For $0 < t < \tau$, the equation becomes $$ \displaystyle x'+\frac{1}{\tau_{\sigma}}x=\frac{F_0}{K_2\tau{\epsilon}}+\frac{F_0\tau_\epsilon}{K_2\tau_\sigma} \delta(t)$$ which has solution $$\displaystyle x(t)=\frac{F_0\tau_{\sigma}}{K_2\tau{\epsilon}}+C_1\exp\left({\frac{t}{\tau_\sigma}}\right)+\frac{F_0\tau_\epsilon}{K_2\tau_\sigma}\Theta(t)\exp\left({\frac{t}{\tau_\sigma}}\right).$$
You can completely avoid going over the Dirac-delta terms.
Your equation has the form $x+ax'=c(F'+bF)$, $x(t)=0=F(t)$ for $t<0$. Now apply the integrating factor $e^{at}$ and compress the right side by a similar factor $$ (e^{at}x(t))'=ce^{(a-b)t}(e^{bt}F(t))' $$ and apply the product rule/partial integration $$ e^{at}x(t)=ce^{at}F(t)-c(a-b)\int_0^te^{as}F(s)\,ds $$ Now use that $$ \int_0^te^{as}Θ(s−τ)\,ds = Θ(t−τ)\frac{e^{a(t-τ)}-1}a $$ to compose the solution \begin{align} x(t)&=cF_0 Θ(t)\left[1+(b-a)\frac{1-e^{-at}}a\right] \\&\quad -cF_0 Θ(t−τ)\left[1+(b-a)\frac{e^{-aτ}-e^{-at}}a\right] \\&\quad +cF_0 Θ(t−2τ)\left[1+(b-a)\frac{e^{-2aτ}-e^{-at}}a\right] \end{align}
Another variation is to observe that $$ (x-cF)'+a(x-cF)=c(b-a)F\implies \Bigl(e^{at}(x-cF)\Bigr)'=c(b-a)e^{at}F $$ which avoids the partial integration step.