I have tried to solve the following problem and got a solution in terms of the complementary error function, but it doesn't satisfy the first boundary condition.
$$\frac{d^2 y(x)}{dx^2} - a y(x) = b \frac{dg(x)}{dx}$$
where
$$g(x) = \delta(x - c), \qquad \lim_{x\to\infty} {y(x)} = 0, \qquad y'(0)=0, \qquad a>0$$
Rephrasing,
$$\ddot x (t) - a \, x (t) = b \, \dot\delta (t - c)$$
Taking the Laplace transform of both sides,
$$\big( s^2 - a \big) \, \hat x (s) = \underbrace{\dot x_0}_{=0} + x_0 s + b \,s \, e^{-cs} = x_0 s + b \,s \, e^{-cs}$$
and, thus,
$$\hat x (s) = x_0 \left(\frac{s}{s^2 - a}\right) + b \left(\frac{s}{s^2 - a} \right) e^{-cs}$$
Taking the inverse Laplace transform,
$$x (t) = \begin{cases} x_0 \cosh \big(\sqrt{a} \, t \big) & \text{ if } 0 \le t \le c^-\\\\ x_0 \cosh \big(\sqrt{a} \, t \big) + b \cosh \big( \sqrt{a} \, (t-c) \big) & \text{ if } t \ge c^+\end{cases}$$
Note that $x$ is discontinuous, as it "jumps" $b$ at time $t=c$.