We have two differential equations, $$\begin{cases} {dT\over dt} = -\alpha(T-B)\\ {dB\over dt} = -\beta(B-T)\end{cases}$$
If $T(0) = 7$ and $B(0) =3$, determine the equilibrium temperature of the system. Is the equilibrium temperature closer to the initial $T$ or initial $B$ value? Explain why this makes sense.
If $t(0) = T(o)$ and $B(0) =B(o)$ determine the eq. temperatur eof the system.
Previously we performed an eigenvalue analysis (assuming $\alpha=1$ and $\beta=2$) to determine the systems behavior. However, Im confused how to do the question asked above, do we just plug the values of those into our system and that's the answer?
the general solution to $x'=Mx$ is $x=C_1 e^{\lambda _1 t} v_1 + C_2 e^{\lambda _2 t} v_2$
where the lambdas and vs are eigenvalues and eigenvectors of M and the Cs are constants.
in your case $M = \left(\begin{array}{cc} -\alpha & \alpha \\ \beta & -\beta \end{array}\right) $
which has $\lambda_1 = 0, v_1 =\left(\begin{array}{c} 1 \\ 1 \end{array}\right) $ and $\lambda_2 = -(\alpha+\beta), v_2 =\left(\begin{array}{c} -\alpha \\ \beta \end{array}\right) $
so your solution is
$\left(\begin{array}{c} T \\ B \end{array}\right)=C_1 \left(\begin{array}{c} 1 \\ 1 \end{array}\right) + C_2 e^{-(\alpha+\beta) t} \left(\begin{array}{c} -\alpha \\ \beta \end{array}\right)$
use the initial conditions to solve for the constants, $C_1$ will be the final temperature.