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Calculus: splitting variables

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    I can't split the x and y variables in each of these three cases. Would somebody be able to help me get started?
     
  2. jcsd
  3. May 8, 2007 #2

    Hootenanny

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    For (a).(i)

    [tex]\frac{dy}{dx} = \frac{x(y^2+3)}{y}[/tex]

    Divide through by [itex](y^2+3)/y[/itex] yielding;

    [tex]\frac{y}{(y^2+3)}\cdot\frac{dy}{dx} = x[/tex]

    For (a).(ii)

    It may be useful to note that;

    [tex]e^{x-2y} = e^x\cdot e^{-2y}= \frac{e^x}{e^{2y}}[/tex]

    As for (b), I don't think it is seperable.
     
    Last edited: May 8, 2007
  4. May 8, 2007 #3
    Regarding a) i) how would I differentiate y/(y²+3)?

    and for a) ii) that's very helpful

    Thanks!
     
  5. May 8, 2007 #4

    Hootenanny

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    For (a)(i), I would use the quotient rule; but I don't think you mean differentiate, I think you mean integrate. If this is the case it may be useful to note that;

    [tex]y = \frac{1}{2}\cdot\frac{d}{dx}(y^2+3)[/tex]

    and

    [tex]\int\frac{f'(x)}{f(x)} = \ln|f(x)| + C[/tex]
     
    Last edited: May 8, 2007
  6. May 8, 2007 #5
    Silly me, of course I meant integrate! :yuck:

    Thanks again!:tongue2:

    Am I on the right track?
    [​IMG]
     
    Last edited: May 8, 2007
  7. May 8, 2007 #6

    Hootenanny

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    Yeah, you've got it basically correct, however, there are a few things I should say.

    On line (5), you should only include one constant of integration (one one side), otherwise they cancel each other out and your left without a constant of integration!

    Secondly, you should ask yourself is there any requirement for the modulus signs? Can y2+3 ever be negative?

    Thirdly, a somewhat minor point, on line (6) your final term should be 2C, not just C.
     
  8. May 8, 2007 #7
    thanks for pointing that out, similiarly with (ii) should I only introduce the constant on one side (I'm assuming so, as the value (1,0) is given to find C...

    Just double checking! :biggrin:
     
  9. May 8, 2007 #8

    Hootenanny

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    Yes, well actually what your doing is;

    [tex]\int f(y) dy = \int g(x) dx[/tex]

    [tex]F(y) + A = G(x) + B[/tex]

    [tex]F(y) = G(x) + C \hspace{1cm}\text{where }C = B-A[/tex]

    Does that make sense?
     
  10. May 8, 2007 #9
    Should I show likewise for a) i)?
     
  11. May 8, 2007 #10

    Hootenanny

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    No not unless you want to :wink:, it isn't usually nesscary unless explicitly stated.
     
  12. May 8, 2007 #11
    Yeah, I don't see why not :approve:

    Could you please double check my value of c for (ii).

    I basically subbed in (1,0) as you do, but the answer seems a bit left of field, so I guess it's best to double check.

    [​IMG]
     
  13. May 10, 2007 #12
    *bump* can anybody confirm my value of c?
     
  14. May 10, 2007 #13
    actually I double checked, I now think c=-1-ln(2)
     
  15. May 14, 2007 #14
    Couldn't seem to edit my previous post, but my final answer is y=(x-1)/2 which sounds alot better!
     
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