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
$(6x - 3y + 2) \, dx - (2x - y - 1) \, dy = 0$
$z = 2x - y$
$dy = 2\,dx - dz$
$(3z + 2) \, dx - (z - 1)(2\,dx - dz) = 0$
$(z + 4) \, dx + (z - 1) \, dz = 0$
$dx + \dfrac{z - 1}{z + 4} \, dz = 0$
$dx + \left(1 - \dfrac{5}{z + 4} \right) \, dz = 0$
$x - \Big[ z - 5 \ln (z + 4) \Big] = c$
$x + \Big[ (2x - y) - 5 \ln (2x - y + 4) \Big] = c$
$-5 \ln (2x - y + 4) + 3x - y = c$ answer
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