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Solve : ( y^2 - 4 )dx - dy = 0

  1. Apr 1, 2005 #1
    I did the following:

    dy/(y^2 -4 ) = dx

    Integral ( -1/4 /y + 2 + 1/4 / ( y-2) . dy = Integral dx

    -1/4 ln | y + 2 | + 1 /4 ln |y -2 | + C = x

    Multiplying all thru by - 4 we get

    ln | y + 2 | + ln | y - 2 | + 4C = 4x ( Note: 4C is another 'C'which is a bigger C)

    ln | y + 2 / y - 2| + C = 4x

    C = 4x - ln | y + 2 / y -2 |

    If this is correct, can we simplify the solution any further.
     
  2. jcsd
  3. Apr 1, 2005 #2
    You ahve a lot of mismatched parentheses but,

    [tex] \frac{1}{y^2-4} = \frac{1}{(y-2)(y+2)} [/tex]

    You cant separate these two in the denominator.

    I would do this using the substitution y = 2sec(t)
     
  4. Apr 1, 2005 #3
    Why not? Using Partial fractions you could

    A/ something + B / something
     
  5. Apr 1, 2005 #4
    I dont think thats what he did, or if he did he didnt show the work. I can barely read his post.
     
  6. Apr 2, 2005 #5

    dextercioby

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    That "-4" multiplication is incorreclty made.You should have other signs...

    Daniel.
     
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