Check solutions of vector Differential Equations

Question

I have solved the vector ODE: $x\prime = \begin{pmatrix}1& 1 \\ -1 &1 \end{pmatrix}x$

I found an eigenvalue $\lambda=1+i$ and deduced the corresponding eigenvector:

\begin{align} (A-\lambda I)x =& 0 \\ \begin{pmatrix}1-1-i & 1 \\-1& 1-1-i \end{pmatrix}x =& 0 \\ \begin{pmatrix} -i&1\\-1&-i\end{pmatrix}x =&0 \end{align} Which is similar to: $\begin{pmatrix}i&-1\\0&0 \end{pmatrix}x = 0$ By Row Reduction. Take $x_2=1$ as $x_2$ is free. We then have the following equation: \begin{align} &ix_1 - x_2 = 0 \\ \iff& ix_1 = 1 \\ \iff& x_1 = \frac{1}{i} \end{align}

Thus the corresponding eigenvector for $\lambda=1+i$ is: $\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix}$.

My solution should then be: \begin{align} x(t) =& e^{(1+i)t}\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\ =& e^t e^{it}\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\ =& e^t\left(\cos(t) + i\sin(t)\right)\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\ \end{align}

By taking only the real parts we have the general solution:

$\left(c_1e^t\cos(t) + c_2e^t\sin(t)\right)\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix}$

How can I quickly check this is correct? Idealy I would like to use Sage to verify. I think this would be faster than differentiating my solution and checking whether I get the original equation.

Answer 1

Let me work through the other eigenvalue, and see if you can follow the approach.

For $\lambda_2 = 1-i$, we have:

$[A - \lambda_2 I]v_2 = \begin{bmatrix}1 -(1-i) & 1\\-1 & 1-(1-i)\end{bmatrix}v_2 = 0$

The RREF of this is:

$\begin{bmatrix}1 & -i\\0 &0\end{bmatrix}v_2 = 0 \rightarrow v_2 = (i, 1)$

To write the solution, we have:

$\displaystyle x[t] = \begin{bmatrix}x_1[t]\\ x_2[t]\end{bmatrix} = e^{\lambda_2 t}v_2 = e^{(1-i)t}\begin{bmatrix}i\\1\end{bmatrix} = e^te^{-it}\begin{bmatrix}i\\1\end{bmatrix} = e^t(\cos t - i \sin t) \begin{bmatrix}i\\1\end{bmatrix} = e^t\begin{bmatrix} \sin t + i \cos t\\ \cos t -i \sin t \end{bmatrix} = e^t\begin{bmatrix}c_1 \cos t + c_2 \sin t\\ -c_1 \sin t + c_2 \cos t\end{bmatrix}$

Note, I put $c_1$ with the imaginary terms, and $c_2$ with the other terms, but this is totally arbitrary since these are just some constants.

For the validation:

• take $x'[t]$ of that solution we just derived.
• then, take the product $Ax$ and verify that it matches the $x'[t]$ expressions from the previous calculation.

I would recommend emulating this with the other eigenvalue/eigenvector and see if you can get a similar result. Lastly, note that $1/i = -i$ (just multiply by $i/i$).