# General form linear differential equation

Normally, if we have an $n$-dimensional linear system of differential equations, we write $$\dot x=Ax,$$ with $x\in\mathbb R^n$. Now I am wondering why the convention isn't to write $$x=A\dot x.$$ It seems to me we want an experssion of $\dot x_i$ for each $i$ in terms of the $x_i$'s - but why not the other way around?

The reason I'm asking this is because when we have an $n$th degree differential equation, they teach us to write it in a system of the form $\dot x=Ax$. I didn't see why it had to be the case that it has this form.

• It doesn't really matter. However, in the form $\dot{x}=Ax$ it looks more similar to the formula $(e^{ax})'=a(e^{at})$, although even this one can be written as $a^{-1}(e^{at})'=e^{at}$ if you really want. – Hellen Oct 26 '17 at 11:33
• Yea, but we don't always have an invertible $A$, so it seems that we really prefer to have $\dot x$ on the left, and we don't care if it would be possible to have $x$ on the left? – Sha Vuklia Oct 26 '17 at 11:35
• Perhaps consider the case when $A$ is 1-dimensional. So $\dot{x} = a x$. If $a \neq 0$ an equivalent equation would be $x= \frac{1}{a} \dot{x}$. But what about the equation $\dot{x} = 0$? There's no way to write this in your alternative way. – Demophilus Oct 26 '17 at 11:46

If we write the system as $\dot{x}=Ax$ we can directly employ the existence and uniqueness theorem for initial value problems of Picard-Lindelöf (see e.g. https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem).
Let us assume the system would look like $x=A\dot{x}$ and $A$ is not invertible, for example let us pick the identity matrix, but with $A_{n,n}=0$. Then you would have $x_n(t)=0$ for all $t$. Therefore you cannot choose an arbitrary initial value for the $n$-th entry in the vector $x(0)$.
• Yes, exactly. By the way the system $x=A\dot{x}$ would be called a differential algebraic equation (see en.wikipedia.org/wiki/Differential_algebraic_equation). – humanStampedist Oct 26 '17 at 12:20