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Differential equation. seperate variables and solve using partial fractions!

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data

    seperate and solve using partial fractions

    dx/dt=9-4x2, x(0)=0


    3. The attempt at a solution

    rearranging gives dx/(9-4x2) = dt
    factorising denominator in preperation for partial fractions becomes

    dx/(3-2x)(3+2x) then A/(3-2x) + B/(3+2x) dx

    so A(3+2x)+ B(3-2x) = 1

    therefore 3A+3B=1 and 2Ax-2Bx=0 therefore A+B=1/3 and A-B=0 therefore A=1/6 and B=1/6

    so integral becomes

    (1/6)/(3-2x) + (1/6)/(3+2x) dx = 1/6*-1/2ln|3-2x| + 1/6*1/2ln|3+2x|

    and RHS is just t+C

    with x(0)=0 then C = 0

    the Answer is given as x(t)=[3(1-e-12t)]/[2(1+e-12t)]

    but i cant seem to manipulate my answer to get there...
     
  2. jcsd
  3. Aug 2, 2011 #2
    your partial fractions look correct and then when you integrated it, it also looks correct. Just combine the logs and then raise both sides to e and then solve for x .
     
  4. Aug 2, 2011 #3
    With c=0...
    -1/12ln((3-2x)/(3+2x))=t
    Ln((3-2x)/(3+2x))=-12t
    (3-2x)/(3+2x)=e^-12t
    ... ?
    How to gather x terms together? This Is where icant get to the answer given :(
     
  5. Aug 2, 2011 #4
    multiply both side by (3+2x) then get all the x's on one side and then factor out the x
     
  6. Aug 2, 2011 #5
    Got it!
    A case of staring at it for too long and missing the basic method. Just needed reassuring I was doing it right.
    Thanks!
     
  7. Aug 2, 2011 #6
    nice
     
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