I am confused by the solution to part $(\mathrm{c})$ of the following question for which I have typed out the full question and solutions for context:
The equation for a driven, damped harmonic oscillator is $$\frac{d^2y}{dt^2}+2\frac{dy}{dt}+(1+k^2)y=f(t)$$ $(\mathrm{a})$ If the initial conditions are $y=0$ and $\dfrac{dy}{dt}=0$ at $t=0$, show that the Green's function, valid for $t\ge 0$, is $$G(t,T)=\begin{cases}A(T)e^{-t}\cos(kt)+B(T)e^{-t}\sin(kt)\quad\text{for}\quad 0\lt t\lt T \\C(T)e^{-t}\cos(kt)+D(T)e^{-t}\sin(kt)\quad\text{for}\quad t\gt T\end{cases}$$
Part $(\mathrm{a})$ solution:
For $t\lt T$, $$\frac{\partial^2G(t,T)}{\partial t^2}+2\frac{\partial G(t,T)}{\partial t}+(1+k^2)G(t,T)=\delta(t-T)=0$$ With a trial solution $G(t,T)\propto e^{mt}$, we have $$m^2+2m+(1+k^2)=0$$ $$\implies m=\frac{-2\pm\sqrt{4-4(1+k^2)}}{2}=-1\pm\sqrt{-k^2}=-1\pm ik$$ Hence $$\begin{align}G(t,T)&=ae^{-1+ik}+be^{-1-ik}\\&=e^{-t}\Big(a(T)e^{ikt}+b(T)e^{-ikt}\Big)\\&=e^{-t}\Big(A(T)\cos(kt)+B(T)\sin(kt)\Big)\\&=A(T)e^{-t}\cos(kt)+B(T)e^{-t}\sin(kt)\end{align}$$ and similarly for $t\gt T$, with $A\to C$ and $B\to D$.
$(\mathrm{b})$ Show that $\mathrm{A}=\mathrm{B}=0$ and so $G(t,T)=0$ for $t\lt T$.
Part $(\mathrm{b})$ solution:
$G(0,T)=0\implies A=0$,
$\dfrac{\partial G}{\partial t}\bigg|_{t=0}=B(T)\cos(0)=0\implies B(T)=0$.
Hence $G(t,T)=0$ for $t\lt T$
$(\mathrm{c})$ By matching $G(t,T)$ at $t=T$, and requiring $\dfrac{dG}{dt}$ to have a discontinuity of $1$ there, show that, for $t\gt T$ $$G(t,T)=\frac{e^{T-t}}{k}\bigg(\cos(kT)\sin(kt)-\sin(kT)\cos(kt)\bigg)$$
Part $(\mathrm{c})$ solution:
Continuity of of $G$ at $t=T$ gives $$\color{blue}{C(T)e^{-T}\cos(kT)+D(T)e^{-T}\sin(kT)=0}$$ Discontinuity of $1$ in $\dfrac{\partial G(t,T)}{\partial t}$ at $t=T$ gives $\color{red}{[-kC(T)e^{-t}\sin(kt)+kD(T)e^{-t}\cos(kt)]_{t=T}-[C(T)e^{-T}\cos(kT)+D(T)e^{-T}\sin(kT)]_{t=T}=1}$
I understand why the part marked blue is correct as it was explained to me in this previous question but I have no idea about the logic behind the red equation. Specifically I don't understand why subtracting those two terms must be equal to $1$.
I notice that the second term on the LHS of the red equation is actually the blue equation and hence equal to zero. I also know from the same previous question why a first derivative must change by $1$. But what is meant by a "discontinuity of $1$" and why does the red equation take that form?
Put in another way, if I wrote "discontinuity of $7$" instead; What do those three words mean? Does it mean that there is a region on the $x$-axis consisting of $7$ units where the function is not present? Or does it mean that there is a region on the $y$-axis where the function is not there?
Any hints or tips is greatly appreciated, thank you.
Using the product rule and showing all intermediate steps to calculate $\dfrac{\partial G(t,T)}{\partial t}$:
\begin{align*}\frac{\partial G(t,T)}{\partial t}&=-kC(T)e^{-t}\sin(kt)-C(T)e^{-t}\cos(kt)+kD(T)e^{-t}\cos(kt)-D(T)e^{-t}\sin(kt)\\&=kD(T)e^{-t}\cos(kt)-kC(T)e^{-t}\sin(kt)-\Big(C(T)e^{-t}\cos(kt)+D(T)e^{-t}\sin(kt)\Big)\\&=\color{red}{[-kC(T)e^{-t}\sin(kt)+kD(T)e^{-t}\cos(kt)]_{t=T}-[C(T)e^{-T}\cos(kT)+D(T)e^{-T}\sin(kT)]_{t=T}}\\&\color{red}=\color{red}{1}\end{align*}
$\fbox{}$
So basically the red equation took that form (one term subtracting the other) because the product rule was being used and the fact that the function $\dfrac{\partial G(t,T)}{\partial t}$ required a discontinuity of one.
In other words there is a jump by one unit in the positive $y$ direction or in this case a jump of one in the positive $\dfrac{\partial G(t,T)}{\partial t}$ direction.