How to solve a differential equation of this form? Please consider a differential equation of the form:
$$ (ax + by + c) dx + (ex + fy + g) dy = 0 $$
For the special case of $c = g = 0$, then this equation is homogenous and I know how to solve it. Normally, I would solve this equation by setting up the following system of
equations:
\begin{align*}
ax + by + c &= 0 \\
ex + fy + g &= 0 
\end{align*}
Assuming that this set of equations of a unique solution, I know how to solve differential equation of this form. However, I how do I solve a differential equation of this form when the above system of differential equations does not have a unique solution?
I am thinking the correct substitution is:
$$ z = ax + by $$
which gives me:
\begin{align*}
dz &= a\, dx \\
dz &= b\, dy 
\end{align*}
Do I have this right?
 A: This result is adapted from "The Fluxional Calculus. An Elementary Treatise" by Thomas Jephson, 1830.
Given
$$\tag 1 (ax + by + c) dx + (ex + fy + g) dy = 0$$
To approach solving this, we need to be complete by doing case work.
Case 1:
Assume $ax+by+c = v$ and $ex +fy+g = w$.
Taking derivatives
$$a dx + b dy = dv \\ e dx + f dy = dw$$
Using elimination
$$dx = \dfrac{f dv - b dw}{af - be} \\ dy = \dfrac{a dw - e dv}{af - be}$$
Substituting these into $(1)$ and multiplying by $(af - be)$, assuming $(af - be) \ne 0$
$$v(f dv - b dw) + w(a dw - e dv) = 0$$
This can be written as
$$(f v -e w)dv + (a w - b v) dw = 0$$
This is a homogeneous case.
Case 2:
Assume $x = v + m$ and $y = w + n$, then $(1)$ becomes
$$(a v + b w + a m + b n + c)dv + (e v + f w + e m + f n + g)dw = 0$$
Suppose
$$(a m + b n + c) = 0 \\ (e m + fn + g) = 0$$
From this, find the required values for $m$ and $n$.
Now this becomes another homogenous case.
Case 3:
If $b = e$ in $(1)$, we have an Exact Equation and can just integrate to get the result.
Case 4:
If $af = be$, then we are dividing by zero and $m = n =  \infty$ and the exactness fails.
However, this case may be reduced to
$$(a x + b y) dx + (a x + b y)dy + c dx + g dy = 0$$
If we substitute $v = ax + by$ and the resulting $dy$, we will have a Separable Equation.
You might also like to review this handy flowchart.
