Solving $y' = Ay + b$

Question

Consider the system of differential equations of the following form

$y' = Ay + b$

where $A$ is an$n \times n$ real matrix and $b$ is a constant vector.

My question is: Can we find the exact solution of this system?

Answer 1

First, we look at the homogeneous equation:

$$y'_h - Ay_h = 0$$

This is solved by

$$y(t) = \exp\left(At\right)C$$

where $\exp(At)$ is the matrix exponential, which exists for any square matrix.

Now, we assume that we can write our particular solution in the following way:

$$y_p(t) = \exp\left(At\right)C'(t)$$

where $C(t)$ is some vector-valued function that we wish to find.

Plugging $y_p(t)$ into the differential equation, we have:

$$\exp(At)C'(t) + A\exp(At)C(t) - A\exp(At)C(t) = \exp(At)C'(t) = b$$

Solving for $C'(t)$, we have:

$$C'(t) = \exp(-At)b$$

and thus

$$C(t) = y_0 +\left[\int_0^t\exp(-As)ds\right]b = y_0+A^{-1}\left[\exp(-At)-I_n\right]b$$

where $I_n$ is the $n\times n$-unit matrix.

Thus, finally:

$$y(t) = \exp(At)y_0 + A^{-1}(I_n-\exp(At))b$$