Let $Lx\equiv x''+px'+qx$, where $p$ and $q$ are constant, be a differential expression. Suppose $x=\phi(t)$ is a solution to $L\phi=0$ with $\phi(0)=0$, $\phi'(0)=1$. Show that a particular solution to $Lx=f(t)$ is given by $x_{p}(t)=\int_{a}^{t}\phi(t-s)f(s)ds$, where $a$ is any constant. Hint: Use Leibniz rule.
Here I'm taking $Lx$ as a differential operator of $x(t)$. So the first thing I did was to solve the homogeneous equation $Lx=0$. This can be done using the fact that $p$ and $q$ are constant so I take the characteristic equation $m^2+pm+q=0$ and get $m=\frac{-p\pm\sqrt{p^2-4q}}{2}$. The solutions are $x_{1}(t)=e^{m_{1}t}$ and $x_{2}(t)=e^{m_{2}t}$ in case that $m_{1}\neq m_{2}$. Given that $\phi(0)=0$ then the solution can't be of these two ways. So I should have that $m_{1}=m_{2}$ so that the solutions be $x_{1}(t)=e^{mt}$ and $x_{2}(t)=te^{mt}$. Given that $m_{1}=m_{2}$ then $p^2=4q$ and $m=-\frac{p}{2}$. So the solution that satisfies $\phi(0)=0$ is $x_{2}(t)=te^{-\frac{p}{2}t}$ and also satisfies $\phi'(0)=1$. Then I find the Wronskian of $x_{1}(t)=e^{-\frac{p}{2}t}$ and $x_{2}(t)=te^{-\frac{p}{2}t}$ that is $W(t)=e^{-pt}$. Now I don't know how to getwhat I want by using the Leibniz rule. If I write the expression for the particular solution $x_{p}(t)$ using the wronskian I get an expression that I can't make equal to what the problem asks.
You're making this too complicated. You have an integral formula for $x_p$ and you're not doing anything with it. Try computing $x_p'(t)$ and $x_p''(t)$ starting by differentiating the integral: $$ x_p'(t) = \frac{d}{dt}\left(\int_a^t \phi(t-s)f(s)ds\right) $$ and applying the Leibniz rule to the integral as the hint suggests.