Why does using elimination in a system of first order differential equations produce an incorrect result? For example, if I have the system,
$$
y'+y=3x \\
y'-y=x
$$
I could then use elimination to minus the top equation from the bottom one to get,
$$
2y=2x \\
y=x
$$
Which is obviously wrong as then, $1+x=3x$ which is wrong.
So why are you not able to use elimination in solving a system of first order differential equations?
 A: The equivalence is really between these systems of equations
$$
\left\vert
\begin{matrix} 
y' + y = 3x \\
y' - y = x
\end{matrix}
\right\vert
\iff
\left\vert 
\begin{matrix}
y' + y = 3x \\
y = x
\end{matrix}
\right\vert
\iff
\left\vert 
\begin{matrix}
y' - y = x \\
y = x
\end{matrix}
\right\vert
$$
so your error is that you dropped one of the original equations, which leads to a larger set of solutions than it should.
However the above equivalence is holding:
The general solution of $y' + y = 3x$ is $y(x) = c \, e^{-x} + 3x -3$.
The general solution of $y' - y = x$ is $y(x) = c e^x-x-1$.
So the first system has no solution. As does the second system. And as does the third system. They all have the emtpy set as set of solutions.
A: No problem using elimination, but your system is incompatible.
A: What the argument shows is that a differentiable function $y(x)$ can satisfy both equations only at a severely restricted set $S$ of $x$ values: with the possible exception of $x=1/2$, every point in $S$ is isolated.  In particular there is no open interval of $x$'s where both equations can hold.
Conversely, for any $S \subset \mathbb{R}$ whose set of non-isolated points is either $\lbrace 1/2 \rbrace$ or empty, there is a differentiable $y(x)$ satisying both equations for all $x \in S$.  The construction is to set $y(x)=x$ and $y'(x)=2x$ for all $x \in S$ and smoothly interpolate $y$ to the points outside $S$.
The converse shows that no further restriction on $S$ can be derived from the equations.
