Given system without control $\dot{x}=ax.$ Given definiton of function in interval $[t,T]$ as
$$J(t)=\frac{1}{2}S(T)x^2(T)+\frac{1}{2} \int_{t}^{T}qx^2(\tau)d\tau.$$
Its Lyapunov equation is: $$-\dot{s}=2as+q$$
From the equation above, How can we prove and Calculate so that we can get the solution as shown below?
$$s(t)=e^{2a(T-t)}S(T)+ \int_{t}^{T}e^{2a(T-t)}qd\tau$$
We have the linear inhomogeneous differential equation with constant coefficients and external forcing $q(t)$.
$$\dot{s}(t)=-2as(t)-q(t)$$
The solution of the homogeneous differential equation $\dot{s}_\text{h}(t)=-2as_\text{h}(t)$ is given by $s_\text{h}(t) = c_1\exp(-2at)$. Now, we use the variation of parameters to obtain the particular solution
$$\text{Ansatz: }s_\text{p}(t) = c(t)\exp(-2at)$$ $$\text{Differential equation: }\dot{s}_\text{p}(t)=-2as_\text{p}-q(t)$$ $$\dot{s}_\text{p}(t)=-2ac(t)\exp(-2at)-q(t)$$ $$\implies \dot{c}(t)\exp(-2at)-2ac(t)\exp(-2at)=-2ac(t)\exp(-2at)-q(t)$$ $$\implies \dot{c}(t)\exp(-2at)=-q(t)$$ $$\implies \dot{c}(t)=-\exp(2at)q(t)$$ $$\implies c(t) = -\int_{\tau=t_0}^{t}\exp(2a\tau)q(\tau)~d\tau$$ $$\implies s_\text{p}(t) =-\exp(-2at)\int_{\tau=t_0}^{t}\exp(2a\tau)q(\tau)~d\tau.$$
The general solution $s(t)$ to a linear differential equation is given by the superposition of the homogeneous solution $s_\text{h}(t)$ and the particular solution $s_\text{p}(t)$. Hence, we obtain
$$s(t) = s_\text{h}(t) + s_\text{p}(t)$$ $$\implies s(t) =c_1\exp(-2at)-\exp(-2at)\int_{\tau=t_0}^{t}\exp(2a\tau)q(\tau)~d\tau.$$
If we use the initial condition $s(t=t_0)=s_0=c_1\exp(-2at_0) \implies c_1=\exp(2at_0)s_0$. The solution is then given by $$\implies s(t) =\exp\left[-2a(t-t_0)\right]s_0-\exp(-2at)\int_{\tau=t_0}^{t}\exp(2a\tau)q(\tau)~d\tau.$$
It is tedious to always derive this result step by step. It is better to remember the following relationships. If we have a linear time-invariant (coefficients are all constant) system
$$\dot{\boldsymbol{x}}=\boldsymbol{Ax}+\boldsymbol{Bu}, \boldsymbol{x}(t=t_0)=\boldsymbol{x}_0.$$
The general solution to this equation is given by
$$\boldsymbol{x}(t) = \exp\left[(t-t_0)\boldsymbol{A}\right]\boldsymbol{x}_0+\int_{\tau = t_0}^t\exp\left[(\tau-t_0)\boldsymbol{A}\right]\boldsymbol{Bu}(\tau)~d\tau.$$