Simplify convoluted variables for partial solution? Suppose one has
$ax' + by' = cx + dy,$ and I want to solve this in terms of an unknown relationship between $x$ and $y$ that will form a solvable differential equation.
As an example, if one had $x' + y' = x + y,$ one could suggest that if $u = x + y,$ then you'd arrive at
$u' = u$ and $u(t) = u_0e^t.$
Similarly, suppose $ax + by = u(t).$
I would then like to extrapolate the technique from that simpler case to this more complex case to achieve a similar result which I suspect is an exponential or pair of exponentials, possibly derived by assuming a change of coordinates related to $\sinh$ and $\cosh$.
How could I rewrite the right-hand sign in terms of $u$?
And if there is an answer to this, is there a generalization that makes sense in terms of Lie symmetries?
 A: Write your equation
$$ax' + by' = cx + dy, \tag{1}$$
under the form:
$$\forall t \in D, \ \ \ ax'(t)-cx(t)=-by'(t)+dy(t)$$
Therefore, the LHS and the RHS define a common function $f(t)$.
As a consequence, (1) equivalent to the system of 2 first order differential equations:
$$\begin{cases} \ \ \ x'(t)&-&cx(t)&=&f(t)\\
-by'(t)&+&dy(t)&=&f(t)\end{cases}\tag{2}$$
where $f$ can be considered in fact as an (almost) arbitrary function.
Of course, one could go a little further. Let us consider for example the case of the first equation in (2). Multiplying LHS and RHS by the integrating factor $a \exp(-\frac{c}{a}t)$, we can trasform it into
$$[ax(t) \exp(-\frac{c}{a}t)]'=\exp(-\frac{c}{a}t)f(t)$$
Integrating, we get:
$$x(t)=\frac{1}{a}\exp(\frac{c}{a}t)\int_0^T\exp(-\frac{c}{a}T)f(T)dT+C \tag{3}$$
with constant $C=x(0)$ (initial condition).
Of course, we have a similar expression for $y(t)$.
Conclusion: (2) (or even (3)) conveys the large degree of freedom we have for "solving" (1).

Edit: (3) can be written under a convolutive form:
$$x(t)=\frac{1}{a}\int_0^Te^{\alpha(t-T)}f(T)dT+C \ \ \tag{3}$$
As $y(t)$ can be expressed under the same form, we can conclude that the most general solution of (1) is:
$$\begin{cases}x(t)&=&A e^{\alpha t}\star f(t)+C\\y(t)&=&Be^{\beta t}\star f(t)+D\end{cases}$$
with fixed constants $A=\dfrac{1}{a}, B, \alpha, \beta$ (depending upon $a,b,c,d$) and arbitrary function $f$ and constants $C,D$.
Remark: we could as well have worked with Laplace Transform on (2)...
