Problem-Solving Strategy for solving the First-Order Circuit(Using the Differential Equations)
STEP 1. Write the equation for the voltage across the capacitor and the equation for the current through the inductor.
STEP 2. Solve the equations that describe the network.
PROBLEM:
Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.
If there are any errors in solution, please let me know in the comments
SOLUTION:
At $t=0-$, the circuit is in steady state and the capacitor acts like a open circuit. The initial voltage across the capacitor can be found using voltage division.
$v_c(0-)=6\times\frac{12k}{12k+8k}=3.6V$
At $t>0$ we can make two KCL equations.
$C\frac{dv_c(t)}{dt}+\frac{v_c(t)}{12k}-\frac{v_o(t)}{8k}=0$
$\frac{v_o(t)+v_c(t)-12}{4k}+\frac{v_o(t)}{8k}=0$
Combining these equations,
$\frac{dv_o(t)}{dt}+\frac{5}{3}v_o(t)=\frac{20}{3}$
The solution to this equation is
$v_o(t)=Ae^{-\frac{5}{3}t}+4$
Using the initial condition $v_c(0-)=v_c(0+)=3.6V$, we find that the complete solution is
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