I'm trying to solve the following differential equation
$$ y'' + \dfrac{1}{2}(y')^2 = A \delta(x) + B \delta(x-a) + C $$
I tried two times, the first one using Laplace transforms, but I don't really know how to deal with the $(y')^2$ term. I found some papers discussing it, but I couldn't really use them since they overcomplicate my original equation. The second attempt has been to find a particular solution to add to the homogeneous one give by $y=\dfrac{1}{c_1+x}$, but I have no clue for where to start.
Does anyone have an idea?
Thank you all!
Assuming $C\gt 0$, the differential equation can be rescaled by substituting $y=2 u$ and $x=\sqrt{2/C}\; t$, leading to the simpler form $$\ddot{u}(t)+\dot{u}(t)^2=1+\alpha\, \delta(t)+\beta \,\delta(t-\tau) \tag{1} \label{1}$$ with $\alpha = A/\sqrt{2C}$, $\beta=B/\sqrt{2C}$ and $\tau = a\sqrt{C/2}$. Comparing the LHS and the RHS of \eqref{1}, it is obvious that $u(t)$ must be continuous at the critical points $t=0$ and $t=\tau$ and only $\dot{u}(t)$ can have jumps at these points (see also the comments by @Sal above). Therefore, setting $\dot{u}(t)= \dot{w}(t)/w(t)$, leading to the linear differential equation $$ \ddot{w}(t)-w(t)= \alpha\, w(0)\, \delta(t) +\beta \,w(\tau)\, \delta(t-\tau) \tag{2} \label{2}$$should pose no problems, unless $w(t)$ acquires zeros for some unfortunate choice of parameters and/or initial conditions.
Restricting ourselves to the special case $\beta=0$, eq. \eqref{2} is solved by employing the ansatz $$w(t)=(c_1 e^t+c_2e^{-t}) \,\Theta(-t)+(c_3e^t+c_4e^{-t})\, \Theta(t), \tag{3} \label{3}$$ where $\Theta(t)$ is the Heaviside step function with $\Theta^\prime(t)=\delta(t)$. The continuity of $w(t) $ at $t=0$ requires $$c_1+c_2 = c_3+c_4 \tag{4} \label{4}$$ and inserting $\ddot{w}(t)$ into \eqref{2} yields the further condition $$c_3-c_4-c_1+c_2=\alpha (c_1+c_2) \tag{5} \label{5}.$$ As a result, $\dot{u}(t)$ is given by $$\dot{u}(t)= \frac{c_1 e^t -c_2 e^{-t}}{c_1 e^t+c_2e^{-t}} \, \Theta(-t)+\frac{c_3 e^t -c_4 e^{-t}}{c_3e^t+c_4e^{-t}} \, \Theta(t) \tag{6} \label{6}$$ and $u(t)$ is found by integrating \eqref{6}.
In the case $\beta \ne 0$, the ansatz $$w(t) =(c_1e^t+c_2e^{-t}) \, \Theta(-t)+(c_3 e^t+c_4 e^{-t}) \, \Theta(t) \Theta(\tau-t)+ (c_5 e^t+c_6 e^{-t}) \, \Theta(t-\tau) \tag{7} \label{7}$$ is used. The further steps of the calculation (analogously to the previous special case) are left as an exercise.
Note that \eqref{1} can be interpreted as the equation of motion of a particle under the influence of a constant force (the term $1$), a friction term $- \dot{u}^2$ (proportional to the square of the velocity) and two "kicks" at the times $t=0$ and $t=\tau$.