First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0

Theorem
The first order ODE:
 * $(1): \quad \left({2 x + 3 y + 1}\right) \mathrm d x + \left({2 y - 3 x + 5}\right) \mathrm d y = 0$

has the solution:
 * $3 \arctan \left({\dfrac {y + 1} {x - 1} }\right) = \ln \left({\left({y + 1}\right)^2 + \left({x - 1}\right)^2}\right) + C$

Proof
Rewriting $(1)$ as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$

we note that it is in the form:
 * $\dfrac {\mathrm d y} {\mathrm d x} = F \left({\dfrac {a x + b y + c} {d x + e y + f} }\right)$

where:
 * $a e = -4 \ne b d = 9$

Hence we can use:
 * First Order ODE in form $y' = F \left({\dfrac{a x + b y + c} {d x + e y + f} }\right)$

which can be solved by substituting:
 * $x := z - h$
 * $y := w - k$

where:
 * $h = \dfrac {c e - b f} {a e - b d}$
 * $k = \dfrac {a f - c d} {a e - b d}$

So let:
 * $x = z - h$ where $h = \dfrac {\left({1 \times 2}\right) - \left({3 \times 5}\right)} {\left({2 \times 2}\right) - \left({3 \times -3}\right)} = \dfrac {-13} {13} = -1$
 * $y = w - k$ where $k = \dfrac {\left({2 \times 5}\right) - \left({1 \times -3}\right)} {\left({2 \times 2}\right) - \left({3 \times -3}\right)} = \dfrac {13} {13} = 1$.

Then $(1)$ is transformed into:


 * $(2): \quad \dfrac {\mathrm d w} {\mathrm d z} = -\dfrac {2 z + 3 w} {3 z - 2 w}$

Let:
 * $M \left({w, z}\right) = 2 z + 3 w$
 * $N \left({w, z}\right) = 3 z - 2 w$

We have that:


 * $M \left({t w, t z}\right) = 2 t z + 3 t w = t \left({2 z + 3 w}\right) = t M \left({x, y}\right)$
 * $N \left({t w, t z}\right) = 3 t z - 2 t w = t \left({3 z - 2 w}\right) = t N \left({x, y}\right)$

Thus both $M$ and $N$ are homogeneous functions of degree $1$.

Thus by definition $(2)$ is a homogeneous differential equation.

So:

By Solution to Homogeneous Differential Equation:


 * $\displaystyle \ln z = \int \frac {\mathrm d u} {f \left({1, u}\right) - u} + C$

where:
 * $f \left({1, u}\right) = \dfrac {2 + 3 u} {3 - 2 u}$

Hence:

Replacing all the substitutions:
 * $\ln \left({x - 1}\right) = \dfrac 3 2 \arctan \left({\dfrac {y + 1} {x - 1} }\right) - \dfrac 1 2 \ln \left({1 + \left({\dfrac {y + 1} {x - 1} }\right)}\right) + C$

Tidying up and reassigning constants appropriately:
 * $3 \arctan \left({\dfrac {y + 1} {x - 1} }\right) = \ln \left({\left({y + 1}\right)^2 + \left({x - 1}\right)^2}\right) + C$