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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?

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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.

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