Take the system of ODEs in terms of the scalar $z$,
$$v'(z) = A e^z + B\cdot v(z)$$ With the initial condition, $$v'(0) = \mathbf{0}$$
For some $B$ an $N \times N$ matix, and $A$ a $1\times N$ vector, $\mathbf{0}$ is a $1\times N$ vector of $0$'s, and $v(z)$ a $1\times N$ vector. Assume whatever you need for the eigenvalues/vectors of $B$, etc.
My question is: what is the solution $v(z)$ in terms of $A$ and $B$? I know how to do it by diagonalizing, getting eigenvectors, etc. for a particular set of matrices/vectors but would rather have it in terms of matrix exponentials. The homogeneous part of the equation is pretty simple in this case, but I am not sure how to do the inhomogeneous part.
(as an addendum to this question, is there a handbook of ODEs that has good coverage of matrix differenital equations?)
Following through on the answer:It appears that the solution is, $$v(z) = e^{B z}\cdot\int_{0}^{z}e^{-B s}\cdot A e^s ds + e^{B s}\cdot C$$ Using the properties a property of matrix exponentials(not proven, see notes) $$=e^{B z}\cdot\left(\int_{0}^z e^{(I - B)s}ds\right) \cdot A + e^{B s}\cdot C$$ Evaluating the integral, $$=e^{B z}\cdot(I - B)^{-1}\cdot(e^{(I-B)z}-I)\cdot A + e^{B s}\cdot C$$ Reorganizing a little $$v(z)=e^{B z}\cdot\left(C - (I - B)^{-1}\cdot A + (I - B)^{-1}\cdot e^{(I - B)z}\cdot A \right)$$ Take the derivative with the product rule for matrix differentiation, $$ v'(z) = e^{B z}(I - B)^{-1}(I - B) e^{(I - B)z} A + B e^{B z}\left(C - (I - B)^{-1}\cdot A + (I - B)^{-1}\cdot e^{(I - B)z}\cdot A \right) $$ Use $v'(0) = 0$ $$ 0 = A + B(C - (I - B)^{-1}A + (I - B)^{-1} A) $$ Solve for $C$, $$C = -B^{-1} A $$ Finally, substitute back into $v(z)$ $$ v(z) = e^{B z}\left((I - B)^{-1}e^{(I - B)z} - B^{-1} - (I - B)^{-1} \right)A $$ Simplifying a little more, $$ v(z) = e^{B z}(I - B)^{-1}\left(e^{(I - B)z} - B^{-1}\right)A $$ Does that look right?
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$\begingroup$Just for completeness:
Let $v(t) \in \mathbb{R}$ be a differentiable function such that:
$$v'(t) - b \, v(t) = f(t),$$ where $b$ is a constant and $f(t) \in \mathbb{R}$ is a continous function of $t$. The solution of this equation can be obtained in terms of an integrating factor, $I(t) \neq 0$, which satisfies:
$$ \frac{\mathrm{d}}{\mathrm{d} t}(I(t) v(t)) = I(t) f(t).$$ It is relatively simple to see that $$I(t) = \exp\left(- \int b \, \mathrm{d}t\right).$$ Then, from the equation above, it yields:
$$ v(t) = C \frac{1}{I(t)} +\frac{1}{I(t)} \int^t_0 I(t) f(t) \, \mathrm{d} t = C e^{-bt}+e^{-bt} \int^t_0 e^{bt} f(t) \, \mathrm{d} t,$$ where the initial condition remains to be imposed (indeed, note that $v'(t) = b \, v(t) + f(t)$) in order to obtain the value of the constant of integration, $C$.
Note that, in the link I give you, $I(t)$ becomes a matrix, $v$ becomes a vector, $b$ is actually a matrix, $f$ becomes a vector, $1/I$ becomes matrix inversion, $\exp$ becomes matrix exponential... so there's a nice correspondence between the 1D case and the $n$-D case that I hope is useful to you.
Cheers!
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