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I am trying to understand a step in the answer to a differential equation which has initial condition: $i_0 = i(t=0)$

$$\frac{di} i + \frac{di}{1-i} = \beta\langle k\rangle \, dt$$

integrate both sides to obtain

$$\ln i - \ln(1-i) + C = \beta\langle k \rangle t $$

Using the initial condition of $i_0 = i(t=0)$, we get:

$$ i = \frac{i_0 e^{\beta \langle k \rangle t}}{1-i_0 + i_0 e^{\beta \langle k \rangle t}} $$

That is the correct answer. However, I am having difficulty moving from line 2 ie $$\ln i - \ln(1-i) + C = \beta\langle k \rangle t $$

Firstly, I am trying to justify the value of C using the initial condition. If i substitute into the second line, I get:

$$ \ln \left\lvert\frac{i}{1-i}\right\rvert + C = \beta \langle k \rangle(0) $$ so, $$ \ln\left\lvert\frac{i}{1-i}\right\rvert + C = 0 $$ $$ \ln\left\lvert\frac{i}{1-i}\right\rvert + C = 0 $$ I am not even sure if I am on the right track. I would appreciate some suggestions on how to proceed.

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  • $\begingroup$ You are on the right track (although it should be $i_0$ instead of $i$ in the last equation). Hence, $C = -\ln\tfrac{i_0}{1-i_0}$. Putting this into the equation that you have for $i$ gives $\ln\tfrac i{1-i} = \beta t+\ln\tfrac{i_0}{1-i_0}$. Now, use the exponential function on both sides and go on with Donald's answer. $\endgroup$ – amsmath Sep 10 '18 at 23:55
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\begin{eqnarray*} \frac{i}{1-i}= \frac{i_0}{1-i_0} e^{\beta(k)t} \end{eqnarray*} Now multiply both sides by$(1-i)(1-i_0)$ and make $i$ the subject of the formula.

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You have $$\ln\lvert\frac{i_0}{1-i_0}\rvert + C = 0$$

Solve for $C$ and substitute in

$$\ln i - \ln(1-i) + C = \beta\langle k \rangle t$$ to get $$\ln i - \ln(1-i) - \ln\lvert\frac{i_0}{1-i_0}\rvert = \beta\langle k \rangle t$$

Combine the logarithms and solve for $i$ to get the final answer.

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  • $\begingroup$ When I combine the logarithms and apply e to both sides I get $$\frac{i}{1-i}\times \frac{i_{0}}{1-i{0}} = e^{\beta \langle k \rangle t}$$ From there I get: $$\frac{i}{1-i} = \frac{1-i{0}}{i_{0}} e^{\beta \langle k \rangle t}$$ I see where @DonaldSplutterwit has something a bit different. He has: $$\frac{i}{1-i} = \frac{i_{0}}{1-i{0}} e^{\beta \langle k \rangle t}$$. Why is that. $\endgroup$ – niz Sep 11 '18 at 1:04
  • $\begingroup$ you should get $\frac{i}{1-i}\times \frac{1-i_{0}}{i{0}} = e^{\beta \langle k \rangle t}$ $\endgroup$ – Mohammad Riazi-Kermani Sep 11 '18 at 1:10
  • $\begingroup$ thank you, i see that. silly oversight. I appreciate you patience. $\endgroup$ – niz Sep 11 '18 at 1:19

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