How to solve a system of linear equations exactly?

Question

Given a system of equations, $$\frac{dR}{dt}=-aR+bJ, \quad \frac{dJ}{dt}=-aJ+bR,$$ I have to discuss what happens to their love(Romeo and Juliet mathematical modeling exercise) is their caution $a$ is larger than their responsiveness $b$, and if $a$ is smaller than $b$. I've been given the option to solve this system exactly or by method of equilibrium and its stability. I'm choosing to go with solving it exactly since I'm still trying to understand the concept of phase planes and phase portraits. Solving it exactly seems like it would just be a method that I have previously learned but can't figure out which.

Answer 1

Differentiating the first equation, $$ R''=-aR'+bJ'. $$ Substituting from the second equation and then the first equation, $$ R''+aR'=b(-aJ+bR) = b^2 R -a(bJ)=b^2R - a(R'+aR) $$ Hence $$ R''+2aR'+(a^2-b^2)R=0, $$ which has general solution $Ae^{-at}\cosh{bt} + Be^{-at} \sinh{bt}$ Since the equations are linear, you can insert this into the $J$ equation to find the corresponding solutions for $J$: $$ J'+aJ = be^{-at}(A\cosh{bt}+B\sinh{bt}). $$ Multiplying by $e^{at}$, $$ (e^{at}J)' = b(A\cosh{bt}+B\sinh{bt}), \\ e^{at}J = A\sinh{bt}+B\cosh{bt}) + C, $$ so $J = e^{-at}A\sinh{bt}+B\cosh{bt}) + Ce^{-at}$. We can see that $C$ is zero from $R'=-aR+bJ$.