Alright, let's work through this problem step by step. First, let's recap the key formula that makes the matrix exponential method so straightforward for linear homogeneous ODE systems:

For a system $\frac{d\mathbf{u}}{dt} = A\mathbf{u}$ with initial condition $\mathbf{u}(0) = \mathbf{u}_0$, the unique solution is given by:
$$\mathbf{u}(t) = e^{tA} \mathbf{u}(0)$$

That's the core of it—we just need to compute the product of the matrix exponential $e^{tA}$ and our initial vector. Let's apply this to your specific problem.

Step 1: Restate the Given Problem #

We have the system:
$$\dfrac{d}{dt} \begin{bmatrix}x \ y \ z\end{bmatrix}=\begin{bmatrix}2 & 2 & -2 \ 5 & 1 & -3 \ 1 & 5 & -3\end{bmatrix} \begin{bmatrix}x \ y \ z\end{bmatrix}$$
with initial condition at $t=0$:
$$\begin{bmatrix}x \ y \ z\end{bmatrix}= \begin{bmatrix}1 \ 1 \ 1\end{bmatrix}$$

And we're told the matrix exponential can be written as:
$$e^{tA}=M{-1}\begin{bmatrix}1&t&\frac{t^2}2\0&1&t\0&0&1\end{bmatrix}M$$

This form tells us matrix $A$ is similar to a Jordan block for the eigenvalue $\lambda=0$ (a triple root, since the Jordan block is 3x3). The matrix $\begin{bmatrix}1&t&\frac{t^2}2\0&1&t\0&0&1\end{bmatrix}$ is the exponential of that Jordan block, denoted $e^{tJ}$.

Step 2: Compute the Transformed Initial Vector #

Let $\mathbf{c} = M^{-1}\mathbf{u}(0)$. This means $M\mathbf{c} = \mathbf{u}(0)$, so we can solve this linear system to find $\mathbf{c}$.

First, we find $M$ (the matrix of generalized eigenvectors for $A$):

• $M$'s first column $\mathbf{v}_1$ satisfies $A\mathbf{v}_1 = 0$ (eigenvector for $\lambda=0$)
• Second column $\mathbf{v}_2$ satisfies $A\mathbf{v}_2 = \mathbf{v}_1$ (generalized eigenvector)
• Third column $\mathbf{v}_3$ satisfies $A\mathbf{v}_3 = \mathbf{v}_2$ (another generalized eigenvector)

Solving these, we get a valid $M$:
$$M = \begin{bmatrix}1&\frac{3}{4}&-\frac{1}{16}\1&1&0\2&\frac{5}{4}&-\frac{7}{16}\end{bmatrix}$$

Now solve $M\mathbf{c} = \begin{bmatrix}1\1\1\end{bmatrix}$:

1. From the second row: $c_1 + c_2 = 1 \implies c_1 = 1 - c_2$
2. Substitute into the first row to find $c_3 = -4c_2$
3. Substitute into the third row to solve for $c_2 = -1$, then $c_1=2$, $c_3=4$

So $\mathbf{c} = \begin{bmatrix}2\-1\4\end{bmatrix}$

Step 3: Multiply by the Jordan Block Exponential #

Next, compute $e^{tJ}\mathbf{c}$:
$$e^{tJ}\mathbf{c} = \begin{bmatrix}1&t&\frac{t^2}2\0&1&t\0&0&1\end{bmatrix}\begin{bmatrix}2\-1\4\end{bmatrix} = \begin{bmatrix}2 - t + \frac{t^2}{2} \cdot 4 \ -1 + 4t \ 4\end{bmatrix} = \begin{bmatrix}2t^2 - t + 2 \ 4t - 1 \ 4\end{bmatrix}$$

Step 4: Transform Back to Original Variables #

Finally, multiply by $M$ to get $\mathbf{u}(t) = M(e^{tJ}\mathbf{c})$:
$$\begin{bmatrix}x(t)\y(t)\z(t)\end{bmatrix} = \begin{bmatrix}1&\frac{3}{4}&-\frac{1}{16}\1&1&0\2&\frac{5}{4}&-\frac{7}{16}\end{bmatrix}\begin{bmatrix}2t^2 - t + 2 \ 4t - 1 \ 4\end{bmatrix}$$

Calculating each component:

• $x(t) = 1\cdot(2t^2 - t + 2) + \frac{3}{4}\cdot(4t - 1) - \frac{1}{16}\cdot4 = 2t^2 + 2t + 1$
• $y(t) = 1\cdot(2t^2 - t + 2) + 1\cdot(4t - 1) + 0\cdot4 = 2t^2 + 3t + 1$
• $z(t) = 2\cdot(2t^2 - t + 2) + \frac{5}{4}\cdot(4t - 1) - \frac{7}{16}\cdot4 = 4t^2 + 3t + 1$

Step 5: Verify the Solution #

Let's confirm this works:

1. Initial condition: At $t=0$, $x(0)=1$, $y(0)=1$, $z(0)=1$—matches our given initial vector.
2. ODE system: Take the derivative of each component and substitute into the original system. For example:
   • $x'(t) = 4t + 2$
   • $2x + 2y - 2z = 2(2t^2+2t+1) + 2(2t^2+3t+1) - 2(4t^2+3t+1) = 4t + 2$—matches $x'(t)$.

All components satisfy the original ODEs, so this is the correct solution.

Final solution:
$$\boxed{\begin{bmatrix}x(t)\y(t)\z(t)\end{bmatrix}=\begin{bmatrix}2t^2+2t+1\2t2+3t+1\4t^2+3t+1\end{bmatrix}}$$

内容的提问来源于stack exchange，提问作者Hideki Ryuga