Question: Show that the problems $ax'' + bx' + cx = f(t); x(0) = 0, x'(0) = v_0$ and $ax'' + bx' + cx = f(t) + av_0 \delta(t); x(0) = x'(0) = 0$ have the same solution for $t \gt 0$. Thus the effect of the term $av_0 \delta(t)$ is to supply the initial condition $x'(0) = v_0$.
The first thing I notice is that these two differential equations involve a few initial values, so naturally I approach the problem using the Laplace transform. I begin with the first equation: $$ax'' + bx' + cx = f(t)$$ becomes $$a(s^2 X(s) - v_0) + bsX(s) + cX(s) = F(s) \rightarrow as^2 X(s) - av_0 + bsX(s) + cX(s) = F(s)$$
Solving in terms of $X(s)$ yields $$X(s) = F(s)G(s) + av_0 G(s)$$ where $G(s) = \frac{1}{as^2 + bs +c}$. I then take the inverse Laplace of this equation to get $$x(t) = \int_{0}^{t} f(\tau)g(t - \tau) d\tau + av_0 g(t)$$
I do the same for the second equation, $ax'' + bx' + cx = f(t) + av_0 \delta(t)$, but end up with $$x(t) = \int_{0}^{t} f(\tau)g(t - \tau) d\tau + av_0 u(t - a)g(t - a)$$
From here I am unsure about what to do. I have two distinct values for $x(t)$, yet I am supposed to show that the two have the same solution? Have I made any errors? What do I do now? Also, what is the intuition behind the very last sentence, specifically that "the effect of the term $av_0 \delta(t)$ is to supply the initial condition $x'(0) = v_0$"?
I think there is a bit of an ambiguity in the problem statement. If $ax^{\prime\prime}+bx^{\prime}+cx=f(t)+av_0\delta(t)$ it implies that $x^{\prime}(t)$ has a jump discontinuity of height $\Delta x^{\prime}=\frac{av_0}a=v_0$ at $t=0$. Thus $x^{\prime}(t)$ is undefined at $t=0$ so it makes more sense to talk about what happens at $t=0^-$ just before the jump discontinuity and $t=0^+$ just after.
In the first differential equation, the question of what $x(t)$ looks like for $t<0$ is moot because the initial values of $x(t)$ and $x^{\prime}(t)$ are given. In the second we are going to assume that for $t<0$, $x(t)=0$ and replace $f(t)$ by $f(t)u(t)$ with the unit step function $u(t)$ turning on the driving function at $t=0$. Then $$\begin{align}x^{\prime}(0^+) & =\int_{-\infty}^{0^+}x^{\prime\prime}(t)dt=\frac1a\int_{0^-}^{0^+}\left(-bx^{\prime}(t)-cx(t)+f(t)u(t)+av_0\delta(t)\right)dt\\ & =\frac1a\left(-0-0+0+xv_0\right)=v_0\end{align}$$ and $$x(0^+)=\int_{=\infty}^{0^+}x^{\prime}(t)dt=0$$ because $x^{\prime}(t)$, $x(t)$, and $f(t)$ are bounded near $t=0$. Both systems have the same driving function, $f(t)$, for $t>0$ and they have the same conditions at $t=0^+$, so they must have the same solutions.