I have 2 ode equations



$R_{12}=2R_{21}$ initial conditions are $P_1=0.1, P_2=0.9$

I solved this with numerical solution by using matlab (ODE45 function)

So I know P1 =0.666 and P2 = 0.333

And I tried to solve this with Laplace transform

so the

$P_2^{laplace} = \frac{P_2(0)s+R_{21}(P_2(0)+P_1(0))}{s^2+sR_{12}+sR_{21}}$

however, is there are any other way to solve this ?

The prof. mentioned about the diagonal matrix, but I dont know how to approach...


1 Answer 1


Turn it into a matrix differential equation (note that I've substituted $A=R_{21}$ and $B=R_{12}$): $$ \frac{d}{dt} \left[ \begin{array}{cc} P_1\\ P_2 \end{array}\right] = \left[\begin{array}{c} -A & B\\ A & -B \end{array}\right]\left[ \begin{array}{c} P_1\\ P_2 \end{array}\right]. $$ Then solve for eigenvalues: $0=(A+\lambda)(B+\lambda)-AB$. So $\lambda=0$ and $-(A+B)$. Find the eigenvectors $\overset{\rightarrow}{\xi}_1$ and $\overset{\rightarrow}{\xi}_2$: $$ \overset{\rightarrow}{\xi}_1=\left[ \begin{array}{c} 1\\ -1 \end{array}\right] \quad \text{ and } \quad \overset{\rightarrow}{\xi}_2=\left[ \begin{array}{c} B\\ A \end{array}\right]$$ Construct the solution: $$ \left[ \begin{array}{c} P_1\\ P_2 \end{array}\right]= c_1\overset{\rightarrow}{\xi}_1\exp{\left(-(A+B)t\right)}+c_2\overset{\rightarrow}{\xi}_2. $$ Then you can plug in whatever initial conditions you like and solve for constants $c_1$ and $c_2$. If you like, the solutions can be written separately as $$P_1(t)=c_1e^{-(A+B)t}+c_2B$$ $$P_2(t)=-c_1e^{-(A+B)t}+c_2A$$

Chapter 7 of Boyce and DiPrima goes over the theory of first order linear systems of equations.

However, notice that adding your equations give $P_1'(t)+P_2'(t)=0$, so $P_2(t)=-P_1(t)+C$. This reduces the equations for $P_1$ to $$\frac{d}{dt}P_1=R_{12}(C-P_1)+R_{21}P_1$$ which can be solved using an integrating factor. Then initial values can be plugged in and we can solve for $P_2$ as well.

  • $\begingroup$ wow thank you so much... $\endgroup$
    – eric
    Feb 23, 2015 at 4:52

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