Differential Equations: $(6x-3y+2)dx - (2x-y-1)dy = 0$

Submitted by agentcollins on

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

Tags
Differential Equation, DE

Jhun Vert Tue, 07/12/2016 - 01:13

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

Let
$z = 2x - y$

$dy = 2\,dx - dz$

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

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

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

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

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

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

$5 \ln (2x - y + 4) - x + y = c$
 

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

Jhun Vert Thu, 07/14/2016 - 06:31

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

Let
$z = 2x - y$

$dy = 2\,dx - dz$

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

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

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

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

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

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

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