differential equations with initial condition

Question

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.

Answer 1

\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.

Answer 2

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.