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Elementary differential equations: transformations

  • Thread starter Mdhiggenz
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  • #1
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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 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.

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
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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

 
Last edited:
  • #3
327
1
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
vela
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To put it bluntly: because what you want to do in your mind is wrong. :smile:

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
327
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Awesome thanks man! (: I love it when math makes sense.
 

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