Question 5.17: Figure 5.46 shows a circuit with resistance, capacitance and......

At t= 0, the current in the circuit is,

2i+4{\frac{\mathrm{d}i}{\mathrm{d}t}}=\mathrm{e}^{-2t}        (5.283)

{\frac{1}{4}}i+{\frac{\mathrm{d}i}{\mathrm{d}t}}={\frac{1}{4}}\mathrm{e}^{-2t}       (5.284)

The auxiliary solution is,

i_{a}=k_{1}\mathrm{e}^{-0.25t}       (5.285)

Considering the particular solution i_{p}=A\mathrm{e}^{-2t} and Eq. (5.284) becomes,

\frac{1}{4}A\mathrm{e}^{-2t}+A(-2)\mathrm{e}^{-2t}=\frac{1}{4}\mathrm{e}^{-2t}       (5.286)

A=-{\frac{1}{7}}      (5.287)

The complete solution is,

i(t)=k_{1}\mathrm{e}^{-0.25t}-{\frac{1}{7}}\mathrm{e}^{-2t}      (5.288)

At t = 0, the initial current in an inductor is zero, i.e. i(0)=0 and Eq. (5.288) becomes,

i(0)=0=k_{1}-{\frac{1}{7}}     (5.289)

k_{1}={\frac{1}{7}}     (5.290)

The final expression of the current is,

i(t)=\frac{1}{7}\bigl(\mathrm{e}^{-0.25t}-\mathrm{e}^{-2t}\bigr)\,\mathrm{A}    (5.291)