Differential Equations: (6x−3y+2)dx−(2x−y−1)dy=0
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
$(6x - 3y + 2) \, dx - (2x -
Solution (1)
(6x−3y+2)dx−(2x−y−1)dy=0
z=2x−y
dy=2dx−dz
(3z+2)dx−(z−1)(2dx−dz)=0
(z+4)dx−(z−1)dz=0
dx−z−1z+4dz=0
dx−(1−5z+4)dz=0
x−[z−5ln(z+4)]=c
x−[(2x−y)−5ln(2x−y+4)]=c
5ln(2x−y+4)−x+y=c
Please double check my solution. It is done in a hurry.
The correct answer is 3x-y+c
In reply to $(6x - 3y + 2) \, dx - (2x - by Jhun Vert
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) ?
I stopped at (z-1)dz/(z+4).
In reply to The correct answer is 3x-y+c by agentcollins
Do long division for (z - 1) ÷ (z + 4)
Solution (1)
In reply to $(6x - 3y + 2) \, dx - (2x - by Jhun Vert
Corrections
(6x−3y+2)dx−(2x−y−1)dy=0
z=2x−y
dy=2dx−dz
(3z+2)dx−(z−1)(2dx−dz)=0
(z+4)dx+(z−1)dz=0
dx+z−1z+4dz=0
dx+(1−5z+4)dz=0
x−[z−5ln(z+4)]=c
x+[(2x−y)−5ln(2x−y+4)]=c
−5ln(2x−y+4)+3x−y=c answer