Solution of Matrix differential equation $\textbf{X}'(t)=\textbf{A}\textbf{X}(t)$

630 Views Asked by At

As I know the solution of the equation system is $\textbf{X}=e^{At}\textbf{X}(0)$.

I want to symbolize fundamental set of solution of matrix dif. equation $\textbf{X}'=\textbf{AX}$ where $\textbf{X}_{4x1}$ and $A_{4x4}$ are matrices. So , I need linearly independent solutions of the Dif. Equation.

(If I have specified Matrix $A$, I can easily find the linearly independent solutions by finding eigenvalues and eigenvectors $A$, but now I can' t write since I don' t know $A$. )

I just know that $A$ has four distinct complex eigenvalues $\lambda_1=a i$, $\lambda_2=-a i$, $\lambda_3=b i$, $\lambda_4=-b i$.

1

There are 1 best solutions below

2
On

Since we $A$ has 4 disctinct eigenvalues, we can write $UDU^{-1}=A$ where $D$ is a diagonal matrix containing $ai, -ai, bi, -bi$ and $U$ is a matrix containing the corresponding eigenvectors of $A$. Then we have

$$\exp(At)X(0)=\exp(UDU^{-1})X(0)=U \begin{pmatrix}e^{ait} & & &\\&e^{-ait}&&\\ &&e^{bit}&\\&&&e^{-bit}\end{pmatrix}U^{-1}X(0).$$

Is this closer to what you're looking for?