Help a stranger please. This is for my homework and I'm having a hard time solving these equations. Show the complete solutions and final answers please. It will be a great help. Thank you so much.

This covers Additional Topics on Equations of Order One, Coefficient Linear in Two Variables.

1. (6x-3y+2)dx-(2x-y-1)dy=0

2. (x+2y-1)dx-(2x+y-5)dy=0

Solution (1)$(6x - 3y + 2) \, dx - (2x - y - 1) \, dy = 0$

$z = 2x - y$

$dy = 2\,dx - dz$

$(3z + 2) \, dx - (z - 1)(2\,dx - dz) = 0$

$(z + 4) \, dx - (z - 1) \, dz = 0$

$dx - \dfrac{z - 1}{z + 4} \, dz = 0$

$dx - \left(1 - \dfrac{5}{z + 4} \right) \, dz = 0$

$x - \Big[ z - 5 \ln (z + 4) \Big] = c$

$x - \Big[ (2x - y) - 5 \ln (2x - y + 4) \Big] = c$

$5 \ln (2x - y + 4) - x + y = c$

Please double check my solution. It is done in a hurry.

The correct answer is 3x-y+c=5ln |2x-y+4| , I can't tell what your mistake is. I stopped at (z-1)dz/(z+4). What did you do there? How it became 1 - 5/(z+4) ?

Do long division for (z - 1) ÷ (z + 4)

Corrections

$(6x - 3y + 2) \, dx - (2x - y - 1) \, dy = 0$

$z = 2x - y$

$dy = 2\,dx - dz$

$(3z + 2) \, dx - (z - 1)(2\,dx - dz) = 0$

$(z + 4) \, dx + (z - 1) \, dz = 0$

$dx + \dfrac{z - 1}{z + 4} \, dz = 0$

$dx + \left(1 - \dfrac{5}{z + 4} \right) \, dz = 0$

$x - \Big[ z - 5 \ln (z + 4) \Big] = c$

$x + \Big[ (2x - y) - 5 \ln (2x - y + 4) \Big] = c$

$-5 \ln (2x - y + 4) + 3x - y = c$

answerPlease help me solve

(10x - 4y - 12)dx - (x - 5y +3)dy =0