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Diff EQNot sure what method to use

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data

    In one of my fluids homeworks, I wound up with a DE of the form


    where A, B, and C are known constants. Can I use the integrating factor on this if I divide thorough by the
    leading coefficient A and put it in the form



    I am just not sure how to approach this DE. Any hints are great :smile:
    Last edited: May 25, 2009
  2. jcsd
  3. May 25, 2009 #2


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    This is seperable.
  4. May 25, 2009 #3
    Okay. Is it separable as is? Because I don't see it? Perhaps a substitution is needed?
  5. May 25, 2009 #4


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    Yes, just divide by B√h + C, and multiply by dt. You will get

    [tex] \frac{-A}{B\sqrt{h} + C}dh = dt [/tex]
  6. May 25, 2009 #5
    Wow. I need to brush up on my math, because I still have no idea how you did that. I have no idea how to divide


    by B√h + C

    I wouldn't even know where to start. I am trying now. Do you do it term by term? I realize that this is just Algebra, but I guess mt Engineering classes
    are veering away from all that goodness
  7. May 25, 2009 #6


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    [tex] A\frac{dh}{dt}+B\sqrt{h}+C=0 [/tex]

    [tex] B\sqrt{h}+C = -A\frac{dh}{dt} [/tex]

    Dividing by B√h + C,

    [tex] 1 = \frac{-A}{B\sqrt{h}+C}\frac{dh}{dt} [/tex]

    Multiplying by dt,

    [tex] dt = \frac{-A}{B\sqrt{h} + C}dh [/tex]
  8. May 25, 2009 #7
    Wow :blushing: I had to see it to believe it! Why can't I see these things?!
    I can bang my head all effin' day trying to figure this stuff out and then wham! Some
    one comes along and is like "hey, why don't you just do this?" and in 2 seconds they're done!

    Thanks dx! I am going to go burn something down now :smile:
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