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1st order diff eq question

  1. Jul 15, 2009 #1
    (2x-y)dx+(2y-x)dy=0 y(1)=3

    solve the given initial value problem and determine at least approx where the solution is valid?

    this should be simple right? I got 2y+2x=C and then 2y+2x=8. It doesnt match the answer in the back of the book at all, and I do see how they could have gotten the answer in the back. Is my answer right? The book says y=[x+sqrt(28-3x^2)]/2, abs(x)<sqrt(28/3)? or is the book right, and if so how do you get to their answer?
    Last edited: Jul 15, 2009
  2. jcsd
  3. Jul 15, 2009 #2
    How did you get your answer, and what's that y(1) on the end?
  4. Jul 15, 2009 #3
    its y(1)=3, I just solve the ivp by using the exact method. It seemed simple but it doesnt match the books answer.

    *my bad posted in wrong forum, if a mod wants to move this go ahead
  5. Jul 15, 2009 #4


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    Gold Member

    This should be in the homework forum. How about showing us your work, this is an extremely easy problem since it is already exact.

    Hint you can directly integrate.
  6. Jul 15, 2009 #5


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    If 2y+ 2x= 8, then y= 4- x dy= -dx so (2x-y)dx+ (2y-x)dy= (2x- 4+ x)dx+ (8- 2x- x)(-dx)= (3x- 4)dx- (8- 3x)dx= -12, not 0. Your solution is clearly wrong.

    Now, how did you get that?

    As djeitnstine said, this is an "exact" differential equation that can be integrated relatively easily.
  7. Jul 15, 2009 #6


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    Indeed, not even the solution is incorrect. I worked it out this morning...however even without working it out its clear the solution seems extraneous.

    Edit: both the OP's solution and the one you claim to have found in the book.
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