Question 8.53: After being open for a day, the switch in the circuit of Fig...

At t<0, i(0- )= 0, v_{c}(0- )= 120 v

For t >0, we have the circuit as shown below.

\frac{120-V}{R}=C \frac{d v}{d t}+i \quad \longrightarrow \quad 120=V+R C \frac{d v}{d t}+i R       (1)
But   v_{L}=v=L \frac{d i}{d t}        (2)
Substituting ( 2 ) into (1) yields

120=L \frac{d i}{d t}+R C L \frac{d^{2} i}{d t^{2}}+i R \quad \longrightarrow \quad 120=\frac{1}{4} \frac{d i}{d t}+80 \times \frac{1}{4} \times 10 \times 10^{-3} \frac{d^{2} i}{d t^{2}}+80 i
or
\left(\mathrm{d}^{2} \mathrm{i} / \mathrm{d} \mathrm{t}^{2}\right)+0.125(\mathrm{di} / \mathrm{dt})+400 \mathrm{i}=600