I am having problem on understanding the below solution regarding constant of integration.(adsbygoogle = window.adsbygoogle || []).push({});

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

$$\frac{3}{10} ln(5-i) = t + k $$$$\frac{1}{10} \int\frac{1}{5-i} di = \frac{1}{3}∫dt .........(1)$$

$$\frac{1}{10}ln(5-i) = \frac{t}{3} + K$$

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

$$\frac{3}{10} ∫\frac{1}{5-i} di = ∫dt$$

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.

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# I Integral constant question

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