Let $B(t)$ be a 1-dimensional Brownian motion. Consider the equation $$\ddot{p}(t)+2\alpha\dot{p}(t)+\beta^2 p(t)+a\dot{\delta}(t)+b\delta(t)=\dot{B}(t) + \ddot{B}(t)$$ where $\alpha,\beta,a,b$ are constants. Initial conditions are $p(0)=p_0$, $\dot{p}(0)=p_1$ and $\ddot{p}(0)=p_2$.
I saw that here on the forum there are some questions related to ODE with Dirac Delta, like e.g. How to solve second order ODE with Dirac Delta?.
In my case, however, to complicate things, I also have Brownian motion. How would you approach an equation like this? For example, is it possible to solve it? Is it possible to write its Ito equation?
First of all, this equation makes no sense as the Brownian paths are barely continuous. So different interpretations of this equation to some regular structure might yield different results.
One can formally construct a first-order system that reduces the differentiation order of the "input" "functions".
Set $u=\dot p+aδ-\dot B$, where $θ$ is the unit jump function. Then $$ \dot u=\ddot p+a\dot δ-\ddot B\\~\\ \dot u + 2α[u-aδ+\dot B]+β^2p(t)+bδ(t)=\dot B(t) $$ This can now be interpreted as SDE system with jumps at the origin \begin{align} dp &= u\,dt-a\,d\theta+dB\\ du &= -(2αu+β^2p)\,dt + (2αa-b)\,d\theta + (1-2α)\,dB \end{align} where $θ$ is the unit jump function, with the distributional derivative $\dot θ=δ$.
Shifting more terms into the first substitution will result in different first-order systems that may or may not give different solutions for $p$ for the same input of the Brownian path $B$.