A resistor of 500 Ω and a capacitor of 4.0 mF are connected in series to a source whose voltage…

Step 1. The time constant. $RC = 500\ \Omega \times 0.004\ \text{F} = 2\ \text{s}$.

Step 2. Standard form. $\dfrac{dq}{dt} + \dfrac{q}{2} = \dfrac{6t}{500} = 0.012\,t$.

Step 3. Integrate. The integrating factor is $e^{t/2}$. Then $q\,e^{t/2} = 0.012\displaystyle\int t\,e^{t/2}\,dt = 0.012\left(2t\,e^{t/2} - 4e^{t/2}\right) + K$.

Step 4. Divide by the factor. $q = 0.024\,t - 0.048 + Ke^{-t/2}$.

Step 5. The constant. At $t = 0\ \text{s}$: $0 = -0.048 + K$, so $K = 0.048$. Then $q = 0.024\,t - 0.048 + 0.048\,e^{-t/2}$.

Step 6. Evaluate. At $t = 4\ \text{s}$: $q = 0.096 - 0.048 + 0.048e^{-2} = 0.0545\ \text{C}$. At $t = 10\ \text{s}$: $q = 0.24 - 0.048 + 0.048e^{-5} = 0.1923\ \text{C}$.

What the straight line means. After a few time constants the transient $0.048\,e^{-t/2}$ is gone. The steady part is $q = 0.024\,t - 0.048$, a straight line, which is the source charge $0.024\,t$ minus a constant lag of $0.048\ \text{C}$. The lag equals the time constant times the charging rate: $2\ \text{s} \times 0.024\ \text{C/s} = 0.048\ \text{C}$.