Understanding the Constant of Integration in RL Circuit Equations

[rajeshmarndi]

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.

 

[BvU]

The 1/3 in (1) should also apply to K ...

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

Shouldn't be the case. Can you show ?

 

[rajeshmarndi]

The 1/3 in (1) should also apply to K ...
Shouldn't be the case. Can you show ?

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,
$$-\frac{1}{10}ln5 = \frac{K}{3}$$
$$∴K = -\frac{3}{10}ln 5 $$​

 

[BvU]

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