Question

A system of differential equations is defined as \frac{dx}{dt} = 7x - 4y \frac{dy}{dt} = 12x - 7y, where $x(0) = 1$ and $y(0) = k$. If $\underline{y}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$, find the solution to this system of differential equations in terms of $k$. Hint: The characteristic polynomial of $\begin{bmatrix} 7 & -4 \\ 12 & -7 \end{bmatrix}$ is $p(\lambda) = \lambda^2 - 1$. $\underline{y}(t) = \begin{bmatrix} \\ \\ \end{bmatrix} + \begin{bmatrix} \\ \\ \end{bmatrix}$ Find a value for $k$ such that $\lim_{t \to \infty} (\underline{y}(t)) = \underline{0}$. k =

          A system of differential equations is defined as
\frac{dx}{dt} = 7x - 4y
\frac{dy}{dt} = 12x - 7y,
where $x(0) = 1$ and $y(0) = k$.
If $\underline{y}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$, find the solution to this system of differential equations in terms of $k$.
Hint: The characteristic polynomial of $\begin{bmatrix} 7 & -4 \\ 12 & -7 \end{bmatrix}$ is $p(\lambda) = \lambda^2 - 1$.
$\underline{y}(t) = \begin{bmatrix} \\ \\ \end{bmatrix} + \begin{bmatrix} \\ \\ \end{bmatrix}$
Find a value for $k$ such that $\lim_{t \to \infty} (\underline{y}(t)) = \underline{0}$.
k =

A system of differential equations is defined as
(dx)/(dt) = 7x - 4y
(dy)/(dt) = 12x - 7y,
where x(0) = 1 and y(0) = k.
If y(t) =
< b m a t r i x >, find the solution to this system of differential equations in terms of k.
Hint: The characteristic polynomial of < b m a t r i x > is p(λ) = λ^2 - 1.
y(t) =
< b m a t r i x >
+
< b m a t r i x >
Find a value for k such that limt →∞ (y(t)) = 0.
k =

Added by Cynthia L.

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