$$ \begin{align} \frac{dI_n}{dt} &= 2(1-p) I_n - I_n + 2(1-p) I_v \\ \frac{dI_v}{dt} &= 2p I_v - 3 I_v + 2p I_n \end{align} $$
I tried to find the eigenvalues and the eigenvectors for this matrix where matrix A is:
$$ A = \left[ \begin{matrix} 1-p & 2(1-p) \\ 2p & 2p-3 \end{matrix} \right] $$
From this matrix, I ended up with the characteristic equation of
$$λ^2+λ(2-p)+2p^2+p-3=0$$
When I used the quadratic equation to solve for the eigenvalues, it becomes too complex an I don't know how to get a "clean" value.
Is there another way of solving for the general solution? Am I on the right track?
mvw is correct, the matrix $A$ should be.
$$ A=\begin{bmatrix} 1-2p&2-2p\\ 2p&2p-3 \end{bmatrix} $$
$A$ has eigenvalues $\lambda_1=-1+2\sqrt{1-p}$ and $\lambda_2=-1-2\sqrt{1-p}$. With corresponding eigenvectors: $$ {\bf v_1}=\begin{bmatrix} -\frac{2(p-1)}{-2+2\sqrt{1-p}+2p}\\ 1 \end{bmatrix} $$ and $$ {\bf v_2}=\begin{bmatrix} -\frac{2(p-1)}{-2-2\sqrt{1-p}+2p}\\ 1 \end{bmatrix} $$
With general solution (in eigenvalues are real and distinct):
$$ \begin{bmatrix} I_n(t)\\ I_v(t) \end{bmatrix} =c_1e^{\lambda_1t}{\bf v_1}+c_2e^{\lambda_2t}{\bf v_2} $$
To move forward from here we need more info like an initial condition and some idea of the value of $p$. If $p=1$ you will not get two linearly independent eigenvectors from $\lambda=-1$, if $p>1$ you will have complex eigenvalues...
Also note that the eigenvalues and eigenvectors were computed with maple. I have not verified them.