Vector $ y = \begin{bmatrix} R(t) & F(t) \end{bmatrix}^{T}$ describes the populations of some rabbits $R(t)$ and foxes $F(t)$. The populations obey the system of differential equations given by $ \dot y = Ay$, where $$ A = \begin{bmatrix} 98 & -1170 \\ 7 & -83 \end{bmatrix}$$ The rabbit population begins at $53,300$. If we want the rabbit population to grow as a simple exponential of the form $R(t) = R_0 e^{8t}$ with no other terms, how many foxes are needed at time $t = 0$?
(Note that the eigenvalues of $A$ are $\lambda = 8$ and $7$.)
Can anyone help me solve this I have no idea where to start? I was thinking of finding the eigenvectors but I'm really sure that would help with the problem. My first initial thought was just to equate $R(t)$ with the initial population but that is obviously wrong, so if anyone can help out I'd be so grateful!
The eigenvalues of $A$ are $\lambda_1=8$ and $\lambda_2=7$, with corresponding eigenvectors $v_1=(13,1)$ and $v_2=(90,7)$.
This means that if you initially have a $13$ to $1$ ratio of rabbits over foxes (i.e. $13$ rabbits per fox), then the application of $A$ will amount to a simple multiplication by $8$ of the corresponding vector.
In math, this means that if $w=\alpha\begin{pmatrix}13\\1\end{pmatrix}$ for any scalar $\alpha$, then $Aw=8w$. This is because any such $w$ is an eigenvector of $A$, and thus $A$ acts on it by producing a multiple of the same vector.
If this is the case, then your system simplifies to $w'=Aw=8w$, which has solution $w(t)=e^{8t} w(0)$.
If your rabbit population must initially be some $R_0$ (equal in your case to $53300$), then this imposes the condition $w(0)=(R_0, R_0/13)=(53300, 4100)$.