Dirac delta on differential equation Lets have the next differential equation:
$$\frac{dx}{dt}=f(t,x)+\sum_{n}g(t,x)\delta (t-t_n)$$
I have to prove that when $t\in (t_i,t_{i+1})$ for some $i$ then:
 $$\frac{dx}{dt}=f(t,x)$$
For this,  I have to use the next property:
$$\int_{a}^b \! f(x)\delta (x-x_0)dx=\left\{ \begin{array}{lcc}
             f(x_0) &   si  & t_1 < x_0 < t_2 \\
             \\0  &  si & x_0 < t_1 \text{ ó } x_0> t_2 \\
            \end{array}\right.$$
Note: I have tried to make the next reassoning:
$$\frac{dx}{dt}=f(t,x)+\sum_{n}g(t,x)\delta (t-t_n) \Rightarrow x(t)= \int_{0}^t a(s,x)ds + \int_{0}^t \sum_{n}b(s,x^-)\delta (s-t_n)ds$$
So, I have to prove that $$\int_{0}^t \sum_{n}b(s,x^-)\delta (s-t_n)ds=0$$ which is not trivial for me :(
 A: We have for any $\epsilon>0$, any fixed $i$, and any smooth test function $\phi$
$$\begin{align}
\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\phi(t)\frac{dx(t)}{dt}dt&=\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\phi(t)F(x(t),t)dt+\sum_{k=1}^n\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\phi(t)g(t,x)\delta(t-t_k)dt\\\\
&=\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\phi(t)F(x(t),t)dt
\end{align}$$
since the Dirac Delta "functions" were not "active" in the interval of integration.  Thus, 
$$\begin{align}
\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\phi(t)\left(\frac{dx(t)}{dt}-F(x(t),t)\right)dt&=0 \tag 1
\end{align}$$
for all test functions $\phi$.  Noting that $(1)$ holds for $\phi(t)=\frac{dx(t)}{dt}-F(x(t),t)$ reveals that 
$$\begin{align}
\int_{t_i+\epsilon}^{t_{i+1}-\epsilon}\left(\frac{dx(t)}{dt}-F(x(t),t)\right)^2dt&=0 \tag 2
\end{align}$$
The only way for $(2)$ to be zero is if the integrand is zero almost everywhere.  Inasmuch as we assumed that $\phi=x'-F$ is a "smooth" function, then we have $\frac{dx}{dt}=F$ as was to be shown.
