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Simple Question

  1. Jun 28, 2009 #1
    1. The problem statement, all variables and given/known data

    scan00031.jpg

    For reference, the bit cut on the side reads, "...to part A, to find the solution of the differential equation."

    3. The attempt at a solution

    The trouble is that i don't know what it's asking me to do in part B, i've completed part A getting 4 / (2x - 3) - 3 / (x + 1). Now does it want me to equal this to y? It says if x = 2 y = 4 so the equation for y would be y = 4 / (2x - 3) - 3 / (x + 1) + 1 . However, what would I do from here, we haven't tackled these types of questions before in class and i've no real idea how to go about answering it.


    Thanks in advance for any help.

    Edit : i've also tried just multiplying out using y but it got nasty and i've a feeling it's quite a simple answer.
     
  2. jcsd
  3. Jun 28, 2009 #2
    The differential equation is separable. Take the equation in part b) and divide both sides by y and multiply both sides by dx. Then you have to integrate both sides, which of course is easier since you did all the work for the right hand side in a).

    EDIT: Here is a link for separable DE's if you've never seen them before:

    http://www.sosmath.com/diffeq/first/separable/separable.html
     
  4. Jun 28, 2009 #3
    Hmmm, that's clever :eek: thanks alot, however it's (x + 1) in the first example and (x - 1) in the second example, which makes it alot more nasty :/, reckon that's a typo?
     
  5. Jun 28, 2009 #4
    Oh sorry, did not notice that. Anyways, just to be safe, I would do the partial fraction decomposition for the expression in b) (not hard) and then integrate. That way, there is no reason for you to get marked off.
     
  6. Jun 28, 2009 #5
    thanks alot for the help
     
  7. Jun 28, 2009 #6

    nrqed

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    You cannot say y = 4 / (2x - 3) - 3 / (x + 1) + 1. I am not sure how you got that.

    Do as the other poster suggested. Divide both sides of the equation by y, multiply by dx and then integrate both sides. You will get y = bunch of logs plus a constant of integration. Now use the value provided y(2)=4 to fix the constant of integration
     
  8. Jun 28, 2009 #7
    [tex]\frac{13-2x}{(2x-3)(x+1)}=\frac{A}{2x-3} + \frac{B}{x+1}[/tex]

    and solve for A and B. Then just plug them back in the first equation and you got the partial fractions.

    Regards.
     
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