Linear Algebra with Differential Equations/Heterogeneous Linear Differential Equations/Diagonalization

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First of all (and kind of obvious suggested by the title), \mathbf{A} must be diagonalizable. Second, the eigenvalues and eigenvectors of \mathbf{A} are found, and form the matrix \mathbf{T} which is an augemented matrix of eigenvectors, and \mathbf{D} which is a matrix consisting of the corresponding eigenvalues on the main diagonal in the same column as their corresponding eigenvectors. Then with our central problem:

\mathbf{X}' = \mathbf{AX} + \mathbf{G}(t)

We substitute:

\mathbf{TY}' = \mathbf{ATY} + \mathbf{G}(t)

Then left multiply by \mathbf{T}^{-1}

\mathbf{Y}' = \mathbf{T}^{-1}\mathbf{ATY} + \mathbf{T}^{-1}\mathbf{G}(t)

As a consequence of Linear Algebra we take the following identity:

\mathbf{D} = \mathbf{T}^{-1}\mathbf{AT}

Thus:

\mathbf{Y}' = \mathbf{DY} + \mathbf{T}^{-1}\mathbf{G}(t)

And because of the nature of the diagonal the problem is a series of one-dimensional normal differential equations which can be solved for \mathbf{Y} and used to find out \mathbf{X}.