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Initial Value Problem (done but i think its wrong please check workthanks)

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    R(dQ/dt) + (1/C)Q = E_0 e^-t ..........Q(0) = 0 and E_0 = a constant


    2. Relevant equations



    3. The attempt at a solution


    first i rearranged to give:
    Q' + (1/CR)Q = (E_0e^-t)/R

    next i multiplied all by integrating factor of: u(t) = e^integ:(1/CR) = e^(t/CR)

    (e^(t/CR) Q)' = (E_0e^-t)/R (e^(t/CR))

    e^(t/CR) Q = integ: (E_0e^-t)/R (e^(t/CR))

    now integrating right side to give...
    e^(t/CR) Q = (E_0/R)e^-t) (e^(t/CR) / (1/CR-1) + C_1

    now rearrange for gen solution:

    Q = (E_0/R)e^-t) / (1/CR-1) + C_1/e^(t/CR)

    then i applied initial conditions to get C_1. The initial condion is: Q(0) = 0

    C = - E_0/R / (1/CR-1)

    so solution is:

    Q = (E_0/R)e^-t) / (1/CR-1) - E_0/R / (1/CR-1) /e^(t/CR)


    is this correct? It doesnt match the solution on exam but not sure if its just cuz i can rearrange it another way..thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 5, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct, but use negative exponent in the second term instead of a fraction to avoid confusion.

    [tex]Q=\frac{E_0}{R}e^{-t}-\frac{E_0}{R(\frac{1}{CR}-1)} e^{-\frac{t}{CR}}[/tex]

    ehild
     
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