Here is what is given:
$$ x' = \begin{bmatrix}1&0 \\2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$
As the title says, we need to find a basis for the set of solutions of this differential equation.
Here is my attempt:
I set up this system
$$\begin{cases} x_1' = x_1 \\ x_2' = 2x_1 + x_2 \end{cases}$$
I then assumed these substitutions were made
$y = x_1$ and therefore $y' = x_2$
From this, I saw that $y'' = 2y + y'$.
Therefore:
$$y'' – y' – 2y = 0$$
I solved this characteristic equation and got these:
$$\begin{cases} y_1 = e^{-t} \\ y_2 = e^{2t} \end{cases}$$
So for my basis I got these:
$$\hat {x^1} = \begin{bmatrix}e^{-t}\\-e^{-t}\end{bmatrix}$$
And:
$$\hat {x^2} = \begin{bmatrix}e^{2t} \\ 2e^{2t} \end{bmatrix}$$
I then selected $t_0 = 0$ as a convenient value of t and then took the determinant of this matrix:
$$\begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}$$
The determinant is equal to $3$ which means that $\hat {x^1}$ and $\hat {x^2}$ are linearly independent and therefore form a basis for the given differential equation.
However, the answer in the back of my book is different (I can post it if anyone needs it). Is my method the wrong way to approach this problem? Did I make a logical or mathematical error? Any help is appreciated.
Best Answer
Sounds like you haven’t covered solving this system of equations directly from the matrix form yet, but I’ll offer this solution for reference.
The solution to the equation $\mathbf x'(t)=A\mathbf x(t)$ is, not surprisingly, $e^{tA}C$, where $C$ is a constant vector that depends on the initial conditions. The exponential of a matrix is defined via a power series, but there are well-known methods for computing it more directly.
We begin by finding the eigenvalues of $A=\small{\begin{bmatrix}1&0\\2&1\end{bmatrix}}$. This matrix is triangular, so we know at a glance that it has the repeated eigenvalue $\lambda=1$. Therefore, we write $A$ as the sum of a diagonal and nilpotent matrix $A=\lambda I+N=I+N$ so that $e^{tA}=e^{t(I+N)}=e^{tI}e^{tN}$. (The last equality is not true for matrix exponentials in general, but holds when the matrices commute.) For the first factor, $e^{tI}=e^tI$. For the second factor $e^{tN}$, we have $N=\small{\begin{bmatrix}0&0\\2&0\end{bmatrix}}$. $N^2=0$, which means that $e^{tN}=I+tN=\small{\begin{bmatrix}1&0\\2t&1\end{bmatrix}}$. Putting the two together, $$e^{tA}=e^t\begin{bmatrix}1&0\\2t&1\end{bmatrix}$$ from which we read our basis $(e^t,2te^t)^T$ and $(0,e^t)^T$.