Question

Consider the differential equation \(\dot{\vec{w}}=A\vec{w}\), with \(\vec{w}(t=0)=\begin{bmatrix}1\\1\end{bmatrix}\). If \(\vec{w}(t)=e^{t}\vec{u}_x+e^{-2t}\vec{u}_y\) is the solution to the equation, where \(\vec{u}_x\) and \(\vec{u}_y\) are unit vectors along the positive \(x\) and \(y\) axes respectively, then which of the following options is the correct matrix representing \(A\)?

• A. \(\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix}\)
• B. \(\begin{bmatrix}-1 & 0\\0 & 2\end{bmatrix}\)
• C. \(\begin{bmatrix}0 & -2\\1 & 0\end{bmatrix}\)
• D. \(\begin{bmatrix}0 & 2\\-1 & 0\end{bmatrix}\)

Hint:

Since u_x and u_y are the coordinate unit vectors, write w(t) as the column vector [e^t, e^{-2t}] and differentiate it directly.

Answer 1

The Correct Option is A

Step 1: Write the solution as a column vector.
Since \(\vec{u}_x\) and \(\vec{u}_y\) are the unit vectors along the \(x\) and \(y\) axes, the given solution can be written as
\[ \vec{w}(t)=\begin{bmatrix}e^{t}\\e^{-2t}\end{bmatrix} \]

Step 2: Check the initial condition.
At \(t=0\), \(\vec{w}(0)=\begin{bmatrix}e^{0}\\e^{0}\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}\), which matches the given initial condition, so this form is consistent.

Step 3: Differentiate the solution component by component.
\[ \dot{\vec{w}}(t)=\begin{bmatrix}\dfrac{d}{dt}e^{t}\\[4pt]\dfrac{d}{dt}e^{-2t}\end{bmatrix}=\begin{bmatrix}e^{t}\\-2e^{-2t}\end{bmatrix} \]

Step 4: Compare with \(A\vec{w}\).
We need \(A\) such that \(A\begin{bmatrix}e^{t}\\e^{-2t}\end{bmatrix}=\begin{bmatrix}e^{t}\\-2e^{-2t}\end{bmatrix}\). The first row must multiply \(e^{t}\) by \(1\) and \(e^{-2t}\) by \(0\), giving the row \([1\ \ 0]\). The second row must multiply \(e^{t}\) by \(0\) and \(e^{-2t}\) by \(-2\), giving the row \([0\ \ -2]\).

Step 5: Write out A and rule out the other options.
\[ A=\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} \] Option (B) has the signs of \(1\) and \(-2\) swapped, which would give \(w_1(t)=e^{-t}\) and \(w_2(t)=e^{2t}\), the opposite growth behaviour to the one given. Options (C) and (D) are off-diagonal matrices; they mix \(w_1\) and \(w_2\) together and would produce coupled or oscillating solutions, not the two independent exponentials \(e^{t}\) and \(e^{-2t}\) seen here.

Step 6: Final conclusion.
\[ \boxed{\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix}} \]