# Method to solve a system of differential equations

I'm studying systems of linear equations.

I'm now specifically studying systems of linear equations of the 1st order, homogeneous:

$Y' = AY$

$A$ as a constant matrix.

Now I know there are various methods to solve this systems. My professor talked about one of the method that consists on reducing the system into a single equation. Does anyone knows where can I find notes or explanations about how to apply correctly this method? I'm having trouble finding it on the internet. I find a lot of things about the method that uses eigenvalues/eigenvectors but that's a method that my professor doesn't like... Does anyone know a good text that I can look for?

• See this answer here – R.W Mar 25 '17 at 18:13

I'm not entirely sure what your professor is hinting at, perhaps ask him/her to elaborate. I am assuming that $y$ is a vector. A homogenous differential equation of the form \begin{align} y' = Ay \end{align} has a solution of the form \begin{align} e^{At}y(0) \end{align} The procedure is similar to the one-dimensional case. Read more about the matrix exponential and how it is useful in solving DEs.

• Hi Erik! Thanks for your fast reply! That's another method my professor gave us: the matrix exponential. However he gave us the alternative method (the one that I'm asking for) because sometimes it's hard to to go to the matrix exponential method. What he's talking is when for example you have a system of two differential equations with two variables and you isolate one of the variables in one of the equations, derive and then substitute in the other equation. So you get a single differential equation of the 2nd order (it's analogous with the nth order with a system of n equations)... – Granger Obliviate Dec 2 '16 at 23:09

If there is a $b$ such that $\{b,Ab,...,A^{n-1}b\}$ is a basis (that is, if the pair $(A,b)$ is completely controllable), then we can reduce the system to a single differential equation.

In the above basis, $A$ has the form (controllable canonical form) \begin{bmatrix} 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & 1 \\ -a_n & -a_{n-1} & \cdots & -a_2 & -a_1 \end{bmatrix} and the system of differential equations looks like $\dot{x}_1 = x_2,..., \dot{x}_{n-1} = x_n$, $\dot{x}_n = -a_n x_1 - \cdots - a_1 x_n$, and by expanding we get the equation $x_1^{(n)} + \sum_{k=0}^{n-1} a_{n-k}x_1^{(k)} = 0$, which is a $k$th order differential equation in $x_1$.

Such a $b$ exists iff in the Jordan normal form of $A$, there is exactly one Jordan block of $A$ associated with any eigenvalue.

• Really, why the downvote? – copper.hat Dec 3 '16 at 17:28
• I dunno, have an upvote to balance it. – Reinstate Monica Dec 4 '16 at 20:38
• @silvascientist: Thanks, I appreciate the sentiment. I'm not concerned about the point, but am wondering about the why! – copper.hat Dec 4 '16 at 20:45
• Such questions are entirely inscrutable when downvoters downvote and then disappear. Alas, I suspect your comment will not summon any explanation from the downvoter; they won't even be notified. – Reinstate Monica Dec 4 '16 at 21:01
• @silvascientist: That is true, however many folks have replied explaining their sentiment, which is all that I ask :-). – copper.hat Dec 4 '16 at 21:03