First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0

Theorem
The first order ODE:
 * $(1): \quad \left({6 x + 4 y + 3}\right) \mathrm d x + \left({3 x + 2 y + 2}\right) \mathrm d y = 0$

has the solution:
 * $3 x + 2 y + \ln \left({\left({3 x + 2 y}\right)^2}\right) + x = C$

Proof
We put $(1)$ in the form:


 * $\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {6 x + 4 y + 3} {3 x + 2 y + 2}$

Substitute $z = 3 x + 2 y$, which gives:
 * $\dfrac {\mathrm d z} {\mathrm d x} = 3 + 2 \dfrac {\mathrm d y} {\mathrm d x}$

and so:
 * $\dfrac {\mathrm d z} {\mathrm d x} = -2 \dfrac {2 z + 3} {z + 2} + 3$

which simplifies down to:
 * $\dfrac {\mathrm d z} {\mathrm d x} = \dfrac {-z} {z + 2}$

This is separable:


 * $\displaystyle - \int \frac {z + 2} z \, \mathrm d z = \int \, \mathrm d x$

which gives:
 * $- z - 2 \ln z = x + C$

which, after the substitutions, gives:
 * $3 x + 2 y + \ln \left({\left({3 x + 2 y}\right)^2}\right) + x = C$