[Cambridge/H2 Maths 2007] The current $I$ in an electric circuit at time $t$ satisfies the differential equation
$4\frac{\text{d}I}{\text{d}t} = 2 - 3I.$

1. Separate the variables to set up the differential equation for integration.

$4\frac{\text{d}I}{\text{d}t} = 2 - 3I$
$\int \frac{1}{2 - 3I} \, \text{d}I = \int \frac{1}{4} \, \text{d}t$

2. Integrate both sides to find the general solution implicitly.

$-\frac{1}{3} \ln |2 - 3I| = \frac{1}{4}t + C$

3. Use the initial condition to find the constant of integration.

Given $I = 2$ when $t = 0$:
$-\frac{1}{3} \ln |2 - 3(2)| = \frac{1}{4}(0) + C$
$-\frac{1}{3} \ln |-4| = C$
$C = -\frac{1}{3} \ln 4$

4. Substitute $C$ back into the equation and eliminate the logarithm.

$-\frac{1}{3} \ln |2 - 3I| = \frac{1}{4}t - \frac{1}{3} \ln 4$
Multiply the entire equation by $-3$:
$\ln |2 - 3I| = -\frac{3}{4}t + \ln 4$
Take the exponential of both sides:
$|2 - 3I| = e^{-\frac{3}{4}t + \ln 4} = e^{\ln 4}e^{-\frac{3}{4}t}$
$|2 - 3I| = 4e^{-\frac{3}{4}t}$

5. Remove the absolute value and make $I$ the subject.

Since $2 - 3I$ is negative when $I = 2$ ($2 - 6 = -4 < 0$), the expression inside the modulus is negative. Therefore, $|2 - 3I| = -(2 - 3I) = 3I - 2$:
$3I - 2 = 4e^{-\frac{3}{4}t}$
$3I = 2 + 4e^{-\frac{3}{4}t}$
$I = \frac{2}{3} + \frac{4}{3}e^{-\frac{3}{4}t}$

1. Solve the differential equation to find the expression for $I$ (as required from the previous dependent part).

Separating the variables for $4\frac{\text{d}I}{\text{d}t} = 2 - 3I$:
$\int \frac{1}{2-3I} \, \text{d}I = \int \frac{1}{4} \, \text{d}t$
$-\frac{1}{3}\ln|2-3I| = \frac{1}{4}t + C$
$2-3I = Ae^{-\frac{3}{4}t}$
Given $I=2$ when $t=0$:
$2-3(2) = A \implies A = -4$
$I = \frac{2}{3} + \frac{4}{3}e^{-\frac{3}{4}t}$

2. Determine the limit of $I$ as $t$ becomes very large.

As $t \to \infty$, the term $e^{-\frac{3}{4}t} \to 0$.
$I \to \frac{2}{3} + \frac{4}{3}(0) = \frac{2}{3}$
Thus, the current approaches a constant value of $\frac{2}{3}$.