Answer to Question #178630 in Differential Equations for LYRA

Question

(3x + y − 2)dx − (3x + y + 4)dy = 0

Answer 1

$\displaystyle (3x + y − 2)\mathrm{d}x − (3x + y + 4)\mathrm{d}y = 0\\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x + y - 2}{3x + y + 4}\\ \textsf{Substitute}\,\, u = 3x + y - 2\\ \frac{\mathrm{d}u}{\mathrm{d}x} = 3 + \frac{\mathrm{d}y}{\mathrm{d}x}\\ \frac{\mathrm{d}u}{\mathrm{d}x} - 3 = \frac{u}{u + 6}\\ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{u}{u + 6} + 3 = \frac{4u + 18}{u + 6}\\ \frac{u + 6}{4u + 18} \mathrm{d}u = \mathrm{d}x \\ \int \frac{1}{4}\cdot \frac{4u + 18 + 6}{4u + 18} \,\, \mathrm{d}u = \int \mathrm{d}x \\ \int \frac{1}{4} + \frac{3}{2} \cdot\frac{1}{4u + 18} \,\, \mathrm{d}u = \int \mathrm{d}x \\ \frac{u}{4} + \frac{3\ln{(4u + 18)}}{8} = x + C\\ \frac{3x + y - 2}{4} + \frac{3\ln{(4(3x + y - 2) + 18)}}{8} = x + C\\ \frac{3x + y - 2}{4} + \frac{3\ln{(2(6x + 2y + 5))}}{8} = x + C\\$