We have a linear system

$$\frac{dx_1}{dt}= -K_1x_1 +K_{21}x_2$$ $$\frac{dx_2}{dt} = K_{12}x_1-K_2x_2$$

So we have solutions of the form

$$x_1= a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t}$$ $$x_2= b_1e^{-\lambda_1 t} + b_2e^{- \lambda_2 t}$$

What I don't understand is how the notes have solutions of the form

$$- \lambda_1 a_1= -K_{1}a_1+K_{21}b_1$$ $$- \lambda_1 b_1= -K_{12}a_1-K_{2}b_1$$ $$- \lambda_2 a_2= -K_{1}a_2+K_{21}b_2$$ $$- \lambda_1 b_2= -K_{12}a_2+K_{2}b_2$$


Since $$x_1= a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t},$$ after differentiating we have $$\frac{dx_1}{dt}=-\lambda_1 a_1 e^{-\lambda_1 t} - \lambda_2 a_2e^{- \lambda_2 t}.$$ On the other hand, $$\frac{dx_1}{dt}= -K_1x_1 +K_{21}x_2=-K_1(a_1e^{-\lambda_1 t} + a_2e^{- \lambda_2 t})+K_{21}(b_1e^{-\lambda_1 t} + b_2e^{- \lambda_2 t})=(-K_1a_1+K_{21}b_1)e^{-\lambda_1 t}+(-K_1a_2+K_{21}b_2)e^{- \lambda_2 t}.$$

Since coefficients by $e^{\lambda_i t}$ are the same, we have $$- \lambda_1 a_1= -K_{1}a_1+K_{21}b_1,$$ $$- \lambda_2 a_2=-K_1a_2+K_{21}b_2.$$ Similarly we get other two relations after differentiating $x_2$.


Here is an "all-matricial" explanation:

$$K:=\pmatrix{-K_1&K_{21}\\K_{12}&-K_2}, \ \ P:=\pmatrix{a_1&a_2\\b_1&b_2}, \ \ D:=\pmatrix{\lambda_1 & 0 \\ 0 & \lambda_2}.$$

The set of 4 equations you have given can be expressed into a single matrix equation, up to a sign change:

$$\pmatrix{a_1&a_2\\b_1&b_2}\pmatrix{\lambda_1 & 0 \\ 0 & \lambda_2}=\pmatrix{-K_1&K_{21}\\K_{12}&-K_2}\pmatrix{a_1&a_2\\b_1&b_2}.$$

In other words:

$$\tag{1}PD=KP \ \ \iff \ \ D=P^{-1}KP,$$

in which we recognize the classical diagonalization relationship that is currently used in the solution of linear differential systems with constant coefficients.

Recall: Let us write the given differential system under the form $X'=KX$. Using change of variable $X=PY$, it becomes:

$$PY'=KPY \iff Y'= \underbrace{P^{-1}KP}_{D} Y$$

If we have chosen $P$ such that $P^{-1}KP=D$, (see equ. $(1)$) where $D$ is a diagonal matrix, we get two "decorellated" very simple differential equations $x'(t)=\lambda_k x(t) \ \ \Rightarrow \ \ x(t)=C_k e^{\lambda_k t}.$


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