Solution of a differential equation can be written as a sum of its homogenous and particular part: $$y = y_h + y_p,$$ or as a sum of zero-state and zero-input solutions: $$y = y_{ZI} + y_{ZS},$$ where zero-input solution $y_{ZI}$ is the homogenous solution $y_h$ with initial conditions applied, and zero-state solution $y_{ZS}$ is the "full solution" $y_h + y_p$, but with the initial conditions set to zero.
So, $$y = y_h + y_p = y_{ZI} + y_{ZS}.$$
Although it is intuitive, and for $t=0$ we can see the equivalence, I would like to see some real proof that this zero-state/zero-input combination truly holds for every $t$, e.g. is always equal to the "default" solution $y_h + y_p$.
Can someone help me with this?
This idea is really Duhamels principle, let us apply it to the general first order ODE:
$y'(t)+ay(t)=f(t)$, $y(0)=g$
Multiply by the integrating factor $e^{at}$ we get:
$\frac{d}{dt}(y(t)e^{at})=f(t)e^{at}$, integrating with respect to $t$ we get:
$y(t)e^{at}-y(0)=\int_0^tf(\tau)e^{a\tau}d\tau$
$\Rightarrow y(t)=y(0)e^{-at}+e^{-at}\int_0^tf(\tau)e^{a\tau}d\tau$
$=y(0)e^{-at}+\int_0^tf(\tau)e^{-a(t-\tau)}d\tau=y_c+y_p$,
it is now very clear that $y_c(0)=y(0)$, and $y_p(0)=0$
Now let $S(t)=e^{-at}$ and we get:
$y(t)=y(0)S(t)+\int_0^tf(\tau)S(t-\tau)d\tau=y(0)S(t)+f*S$. (Where $*$ denotes the convolution).
Also this still holds for non constant coefficients, I.e. consider:
$y'(t)+a(t)y(t)=f(t)$, $y(0)=g$ here we take $S(t)=e^{-\int_0^ta(\tau)d\tau}$, and we obtain:
$y(t)=y(0)S(t)+f*S$