# Show that $w(t)-Cx(t)=\exp(q(t-t_k))ξ(t_k)+\int_{t_k}^{t}\exp(q(t-s))C(f(z(s),u(s))-f(x(s),u(s))+q(x(s)-z(s))) \,dx$

Show that $$w(t)-Cx(t)=\exp(q(t-t_k))ξ(t_k)+\int_{t_k}^{t} \exp(q(t-s))C(f(z(s),u(s))-f(x(s),u(s))+q(x(s)-z(s))) \,dx$$ combining the equations $$(1) \ y(t_k)=h(x(t_k))+ξ(t_k),(2)\ h(x)=Cx, (3)\ w(t_k)=y(t_k), (4)\ \dot w=\bigtriangledown h(z(t))f(z(t),u(t))-K(z(t),w(t),u(t))(w(t)-h(z(t))), (5)\ \dot z=f(z,u)+g(z,y,u)\cdot (y-h(z))$$ with $$x\in R^n,u\in U, y\in R^p, ξ \in R^p, z\in R^n,w \in R^p,h:R^n\to R^n,g:R^n\times R^p \times U \to R^{n \times p}, f:R^n \times U \to R^n$$ and an increasing sequence $$(t_k,k=1,2,...)$$ with $$t_0=0$$ and $$\lim_{k\to\infty} t_k=\infty$$. Also we can assume that $$Κ(z,w,u)=-qI$$

ATTEMPT From $$(1)$$, $$(3)$$ we have that $$w(t)-h(x(t))=ξ(t)$$. We also know that $$h(x)=Cx$$, so $$w(t)-Cx=ξ(t)$$. We must show that $$ξ(t)=\exp(q(t-t_k))ξ(t_k)+\int_{t_k}^{t} \exp(q(t-s))C(f(z(s),u(s))-f(x(s),u(s))+q(x(s)-z(s))) \,dx$$ We also know that $$\dot w=\bigtriangledown h(z(t))f(z(t),u(t))-K(z(t),w(t),u(t))(w(t)-h(z(t)))$$ So, $$\dot w=C \dot z(t) f(z(t),u(t))+qI(w(t)-h(z(t))\implies \dot w=C \dot z(t) f(z(t),u(t))+qI(w(t)-Cz(t))$$ From $$(5)$$, $$\dot w(t)=C(f(z,u)+g(z,w,u)(w(t)-Cz(t))f(z,u)+qI(w(t)-cz(t))$$

So, from now and then I could not figure out how to proceed so as to reach out the requested equation. I would be very helpful if someone can help me with this. All my attempts reach a dead end.