MHB Differential equation with a matrix

[Rorschach]

Suppose we have the matrix $ \mathbf{N} = \begin{bmatrix} 4 & -2 \\ -2 & 1 \end{bmatrix}$ and $\mathbf{X} = \begin{bmatrix}x \\ y \end{bmatrix}$. I want to solve $\displaystyle \frac{d\mathbf{X}}{dt} = \mathbf{NX}$.

The eigenvalues of the matrix are $\lambda_1, \lambda_2 = 0,5$ and eigenvectors are $v_1 = (1,2)^t$ and $v_2 = (2, 1)^t$.

So I thought the solution would be $\mathbf{X} = c_1 (1,2)^t + c_2(2,1)^{t}e^{5t}$. But apparently this is wrong? Why?

P.S. the t on the vectors denotes transpose, not to be confused with the t on the exponential.

 

[S.G. Janssens]

$v_2$ is not an eigenvector for the eigenvalue $\lambda_2 = 5$. Try it with $(-2, 1)^t$.
The rest seems fine to me.

 

[Rorschach]

$v_2$ is not an eigenvector for the eigenvalue $\lambda_2 = 5$. Try it with $(-2, 1)^t$.
The rest seems fine to me.

Thank you. Is it technically correct to say that $(1, 2)^t$ is a solution to the differential equation? This question was a multiple choice, and the only answer that fits is $(1,2)^t$. I think this is the correct answer because $c_1 (1,2)^t+c_2(-2,1)^te^{5t} = (1,2)^t$ for $c_1 = 1, c_2 = 0$.

 

[S.G. Janssens]

Thank you. Is it technically correct to say that $(1, 2)^t$ is a solution to the differential equation? This question was a multiple choice, and the only answer that fits is $(1,2)^t$. I think this is the correct answer because $c_1 (1,2)^t+c_2(-2,1)^te^{5t} = (1,2)^t$ for $c_1 = 1, c_2 = 0$.

The whole line $\{c_1 (1,2)^t \,:\, c_1 \in \mathbb{R}\}$ consists of constant solutions of the differential equation, and choosing $c_1 = 1$ selects one of them, so you are right indeed.

 

[Rorschach]

Thank you very much!

 

[HOI]

A less "sophisticated" version: The differential equation $\frac{dX}{dt}= NX$, with $N= \begin{bmatrix}4 & -2 \\ -2 & 1\end{bmatrix}$ can be written, by taking $X= \begin{bmatrix}x \\ y \end{bmatrix}$ as $\begin{bmatrix}\frac{dx}{dt} \\ \frac{dy}{dt} \end{bmatrix}= \begin{bmatrix}4x- 2y \\ -2x+ y\end{bmatrix}$.

That is equivalent to the two differential equations
$\frac{dx}{dt}= 4x- 2y$ and $\frac{dy}{dt}= -2x+ y$.

Differentiate the first equation again to get $\frac{d^2x}{dt^2}= 4\frac{dx}{dt}- 2\frac{dy}{dt}$. Using the second equation, we can replace that $\frac{dy}{dt}$ by $-2x+ y$ to get
$\frac{d^2x}{dt^2}= 4\frac{dx}{dt}- 2(-2x+ y)= 4\frac{dx}{dt}+ 4x- 2y$

But from the first equation $-2y= \frac{dx}{dt}- 4x$ so that becomes
$\frac{d^2x}{dt^2}= 4\frac{dx}{dt}+ 4x+ \frac{dx}{dt}- 4x= 5\frac{dx}{dt}$

The characteristic equation for the differential equation $\frac{d^2x}{dt^2}+ 5\frac{dx}{dt}= 0$ is $r^2+ 5r= r(r+ 5)= 0$ which gives r= 0 and r= -5. The general solution is $x(t)= A+ Be^{-5t}$. From the equation $-2y= \frac{dx}{dt}- 4x$, $-2y= -5Be^{-5t}- 4A- 4Be^{-5t}= -4A- 9Be^{-5t}$.

So $X(t)= \begin{bmatrix}A+ Be^{-5t} \\ -4A- 9Be^{-5t} \end{bmatrix}= A\begin{bmatrix}1 \\ -4 \end{bmatrix}+ Be^{-5t}\begin{bmatrix}1 \\ -9 \end{bmatrix}$.