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I Integral constant question

  1. Dec 6, 2017 #1
    I am having problem on understanding the below solution regarding constant of integration.
    On integrating an differential equation of RL circuit , for e.g
    $$10i + 3\frac{di}{dt} = 50 $$
    $$i.e \frac{di}{50-10i} =\frac{dt}{3}$$
    Integrate
    $$\frac{1}{10} \int\frac{1}{5-i} di = \frac{1}{3}∫dt .........(1)$$
    $$\frac{1}{10}ln(5-i) = \frac{t}{3} + K$$
    Since i(0) = 0,
    $$\frac{1}{10}ln(5-0) = 0 + K$$
    ( I understand since it's an RL circuit equation, I assumed when i=0 , also t=0 )
    $$K = \frac{-ln5}{10}$$


    Now if I do the above solution at (1), as
    $$\frac{3}{10} ∫\frac{1}{5-i} di = ∫dt$$​
    $$\frac{3}{10} ln(5-i) = t + k $$
    At i(0)=0,$$K = - \frac{3}{10}ln 5$$

    I get two different value of K.
    On substituting the value of K in the equation. I will get different value of i.

    Please clarify the mistake. Thanks.
     
  2. jcsd
  3. Dec 6, 2017 #2

    BvU

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    The 1/3 in (1) should also apply to K ...
    Shouldn't be the case. Can you show ?
     
  4. Dec 6, 2017 #3
    I was looking the below link, example 8 solution
    https://www.intmath.com/differential-equations/2-separation-variables.php

    This is how it integrate.
    Integrate
    $$\frac{1}{10} \int\frac{1}{5-i} di = \frac{1}{3}∫dt .........(1)$$
    $$-\frac{1}{10}ln(5-i) = \frac{t}{3} + K ........(2) $$
    (I missed the -ve sign in my post)
    Since i(0) = 0,
    $$-\frac{1}{10}ln(5-0) = 0 + K $$
    (-ve sign was also missing here earlier)
    $$K = \frac{-ln5}{10}.......(3)$$

    Does it mean ...(2) is wrong in the above link
    and it should be as,
    $$-\frac{1}{10}ln(5-i) = \frac{1}{3} (t+ K) ......(2)$$
    Hence, since i(0)=0 and t=0,
    $$-\frac{1}{10}ln5 = \frac{K}{3}$$
    $$∴K = -\frac{3}{10}ln 5 $$​
     
    Last edited: Dec 6, 2017
  5. Dec 6, 2017 #4

    BvU

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    Can you see that both expressions for ##\ i(t)\ ## come out the same in you first post (with signs fixed) ?
     
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