Elementary differential equations: transformations

[Mdhiggenz]

Homework Statement

(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 don't 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.

Homework Equations

The Attempt at a Solution

 

[vela]

Homework Statement

(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

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.

What I don't 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.

Homework Equations

The Attempt at a Solution

 

[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?

 

[vela]

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.

 

[Mdhiggenz]

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