Suppose we have two differential equations $$ \frac{dx}{dt} = a \frac{dr}{dt} - b\frac{dR}{dt},$$ and $$ \frac{dy}{dt} = c \frac{dr}{dt} + d\frac{dR}{dt}. $$
My goal here is not to solve the differential equations, only to rearrange it such that I can get something like that
$$ \frac{dy}{dt} = f\left(\frac{dx}{dt}\right), $$ where $f$ is independent $r'$ and $R'$.
The reason for this is because the system corresponds to a physical situation in which I can measure $x'$ and $y'$ but not $R'$ or $r'$., so I would really appreciate the help.
Two attempts I thought of:
- Solving one for $r'$ and plugging into the other. This method of elimination seems to not be possible unless I am missing something.
- It did occur to me that I could write the equation in terms of matrices (matrix). I thought that the eigenvalues and/or eigenvectors could be useful but here I wasn't sure how. $$ \pmatrix{y' \\ x'} = \pmatrix{a & -b \\ c & d}\pmatrix{r' \\ R'}.$$
I have had courses in differential equations, but we never covered anything like this.
$$\begin{cases} \frac{dx}{dt} = a \frac{dr}{dt} - b\frac{dR}{dt}\\ \frac{dy}{dt} = c \frac{dr}{dt} + d\frac{dR}{dt} \end{cases} $$ This is a system of 2 equations with 4 unknown functions : $x(t)$ , $y(t)$ , $r(t)$ and $R(t)$. This means that one can arbitrary chose two functions among the four and determine the two others. For example, chose arbitrary any functions $r(t)$ and $R(t)$, then the solution is : $$\begin{cases} x(t) = \int \left(a \frac{dr}{dt}\right)dt - \int \left(b \frac{dR}{dt}\right)dt \\ y(t) = \int \left(c \frac{dr}{dt}\right)dt + \int \left(d \frac{dR}{dt}\right)dt \end{cases} $$ If the coefficients $a,b,c,d$ are not functions of $t$ : $$\begin{cases} x(t) = a\:r(t)-b\:R(t)\\ y(t) = b\:r(t)+d\:R(t) \end{cases} $$ with any functions $r(t)$ and $R(t)$.
IN ADDITION, after the wording of the question was updated :
As explained above, because they are only two equations for four unknown functions, each one among the unknown functions is related to two of the other ones. This means that $y'$ is related to $x'$ AND either with $r'(t)$ or with $R'(t)$ : $$\begin{cases} y'=-\frac{d}{b}x'+\frac{ad+bc}{b}r'\\ y'=\frac{c}{a}x'+\frac{ad+bc}{a}R' \end{cases}$$ Thus, in the general case, a relationship of the kind $y'=f(x')$ is impossible.
Nevertheless, it is possible in particular case, if a particular relationship exists between $a,b,c,d$ :
If $\quad ad+bc=0\quad\implies\quad y'=k\:x' \quad\text{ with }\quad k=\frac{c}{a}=-\frac{d}{b}$
So, in this particular case, the function $f$ is a simple linear relationship, as expected since all relationships are linear in the whole problem.
Note that this particular case occurs when the determinant of the matrix $ \left|\begin{matrix} a & -b \\ c & d \end{matrix}\right|=0$.