Solving coupled first-order linear ODEs Basically this question comes from population modelling. Let $y$ be the population of Lions and let $x$ be the population of deer. By ignoring the effect of deer, we observe that $$dy/dt = k_2 y$$ [Exponential]. By ignoring the effect of lions on their own population, we obtain that $$dy/dt = k_1 x$$ Similarly, for deer, we observe that ignoring lions, $$dx/dt = k_3 x$$ and ignoring effect of current deer population, $$dx/dt = -k_4 y$$ Note the minus sign as the lions eat the deer. Thus, we can form coupled equations \begin{align} dy/dt &= k_1 x + k_2 y  \tag{1} \\ dx/dt &= k_3 x - k_4 y \tag{2} \end{align}
I have been trying to solve these, but have been having considerable difficulty. If it makes the problem any easier, you can take $$k_i = 1 \quad \forall\ i$$ (Although this is inaccurate in this case, but it will hint at the correct form of solution.) I tried solving this version too, but I am not able to reach a result. Thanks in advance. 
 A: This is a standard two dimensional linear system:
$$\begin{bmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{bmatrix} = \begin{bmatrix} k_3 & -k_4 \\ k_1 & k_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$
If the matrix is denoted by $A$, the solution is $(x(t),y(t))^T = e^{At} (x(0),y(0))^T$. One can compute $e^{At}$ by diagonalizing $A$: if $A=PDP^{-1}$ for a diagonal $D$ then $e^{At} = P e^{Dt} P^{-1}$ where $e^{Dt}$ is a diagonal matrix with $(e^{Dt})_{ii} = e^{D_{ii} t}$. This will be possible provided $(k_3+k_2)^2 - 4 (k_3 k_2 + k_1 k_4) \neq 0$. If it is zero (for example, if $k_3=1$, $k_2=2$, $k_1=k_4=1/2$) then the matrix exponential may be slightly harder to compute, but the procedure can be found in any elementary ODE textbook, such as Boyce and DiPrima or Brannon and Boyce.
By the way, basic predator-prey models usually model the interaction between the predators and the prey via an $xy$ term instead. Essentially, your $k_4$ and $k_1$ should not be constant, because each predator will eat (and thus reproduce) more if there are more prey, while more prey will be eaten if there are more predators. 
Such models have nice properties. For "typical" parameter values, they have a nontrivial equilibrium around which certain trajectories will oscillate. Your system will not do this: any nonzero initial condition will diverge, because the trace is $k_2 + k_3 > 0$ and the determinant is $k_3 k_2 + k_1 k_4 > 0$, so both eigenvalues have positive real part.
A: Hint
Starting with $$\frac{dy}{dt} = k_1 x + k_2 y \tag{1}$$ $$\frac{dx}{dt} = k_3 x - k_4 y \tag{2}$$ there is another possible way. Eliminating $x$ from the first equation gives $$x=\frac{1}{k_1} \frac{dy}{dt}-\frac{k_2}{k_1} y$$ then$$\frac{dx}{dt}= \frac{1}{k_1} \frac{d^2y}{dt^2}-\frac{k_2}{k_1}\frac{dy}{dt}=\frac{k_3}{k_1} \frac{dy}{dt}-\frac{k_2 k_3}{k_1} y-k_4 y$$ which gives a second order differential equation in $y$ $$\frac{d^2y}{dt^2}-(k_2+k_3)\frac{dy}{dt}+(k_1k_4+k_2k_3)y=0$$ Solve it for $y$ and compute from it according to $$x=\frac{1}{k_1} \frac{dy}{dt}-\frac{k_2}{k_1} y$$ 
