# Elementary differential equations: transformations

1. Sep 16, 2012

### Mdhiggenz

1. The problem statement, all variables and given/known data

(x+2y+3)dx+(2x+4y-1)dy=0

a1= 1 b1=2 a2=2 b2=4
a2/a1=b2/b1

Therefore z=x+2y
Here is where I get confused I understand that they must get a dz in the equations thus they take the derivative with respect to y of the following equation z=x+2y
thus giving dz=dx+2 or dz-dx/2

What I dont understand is the simplification process that occurs now

(z+3)dx +(2z-1)(dz-dx/2)=0

then the book goes on to transform it into

7dx + (2z-1)dz=0

I have no idea what they did to get to that step.

2. Relevant equations

3. The attempt at a solution

2. Sep 16, 2012

### vela

Staff Emeritus
This isn't right. When you differentiate, you should get dz = dx + 2 dy. You can't have that lone 2 hanging around. It needs to be multiplied by dy. Then when you solve for dy, you get dy = (dz-dx)/2. Your second mistake was leaving out the parentheses.

Does it make sense after you make those corrections? It's just algebra from here.

Last edited: Sep 16, 2012
3. Sep 16, 2012

### Mdhiggenz

It does clear up a few things but I am actually stuck on the algebra portion.

What I am doing in that case is first multiplying both sides by 2 and I get

2zdx+6dx+(2z-1)(dz-dx)=0
2zdx+6dx+(2zdz-2zdx-dz+dx)=0

7dx+2zdz-dz=0

In my mind I want to make it 7dx+zdz=0

But the book factors out the dz to make it 7dx+(2z-1)dz=0

Why do they go that route?

4. Sep 16, 2012

### vela

Staff Emeritus
To put it bluntly: because what you want to do in your mind is wrong.

When you combine the dz terms, what you are doing is factoring. You pull out the common factor of dz to get (2z-1) dz. You can't do anything to simplify the expression from here.

If you had, instead, 2z dz - z dz, you'd factor dz out to get (2z-z)dz. This time, you can simplify what's in the parentheses to end up with z dz. Or even better, you could pull out the common factor of z dz to get 2z dz - z dz = (2-1) z dz = z dz.

5. Sep 16, 2012

### Mdhiggenz

Awesome thanks man! (: I love it when math makes sense.