What is the solution to the matrix differential equation:
$$ \frac{dx}{dt} = \boldsymbol{A}x(t) + \boldsymbol{B}u$$
Where, $ x(0) = x_0 $, $A_{n \times n}$ is a square matrix and $B_{n\times1}$ is $n\times1$ matrix
Assuming u is constant for the time interval $[0, T]$
Also please refer me to a resource where I can learn to solve this types of equations, because solving the equations were not covered in my linear algebra course.
Best Answer
Define the change of variables $z(t):=\exp(-At)x(t)$. We then obtain
$$\dot{z}(t)=-Az(t)+\exp(-At)\dot{x}(t)$$
and thus
$$\dot{z}(t)=-Az(t)+\exp(-At)(Ax(t)+Bu(t))$$
which yields
$$\dot{z}(t)=\exp(-At)Bu(t),$$
where we have used the fact that $\exp(-At)A=A\exp(-At)$. Integrating from 0 to $t$, we obtain
$$z(t)=z(0)+\int_0^t\exp(-As)Bu(s)ds.$$
Since $z(0)=x_0$ and $x(t)=\exp(At)z(t)$, we obtain
$$x(t)=\exp(At)x_0+\int_0^t\exp(A(t-s))Bu(s)ds.$$
This is the general solution to your equation. Any textbook on (linear) dynamical systems or on state-space methods for control theory will address this.
When the input is constant and the matrix $A$ is invertible, we can arrive to the following expression by simply evaluating the integral
$$x(t)=\exp(At)x_0+A^{-1}(\exp(At)-I)Bu.$$