91Ó°ÊÓ

An RLC circuit, like the one in this problem, is analyzed using second-order differential equations. These equations involve the second derivative of the unknown function, which is, in our case, the charge on the capacitor, denoted as \( q(t) \). The standard form of a second-order linear differential equation is:

• \( a q''(t) + b q'(t) + c q(t) = f(t) \)

Here, \( a \), \( b \), and \( c \) are constants depending on the circuit parameters, and \( f(t) \) is the forcing function representing the external voltage applied. In analyzing these circuits, we often convert the physical quantities like inductance \( L \), resistance \( R \), and capacitance \( C \) into the coefficients of a differential equation. For this exercise, we substitute:\( L = \frac{1}{2} \), \( R = 10 \), and \( C = 0.01 \), leading us to:\[ \frac{1}{2}q''(t) + 10q'(t) + 100q(t) = E(t) \].By solving this equation, we can determine the behavior of the charge and its dynamic responses over time.

Initial value problems require solving differential equations with specified values at the beginning of the period under consideration. Here, the initial conditions are given by \( q(0) = 0 \) and \( q'(0) = 0 \), representing the initial charge and initial current (rate of change of charge), respectively.
These initial conditions are crucial because:

• They allow us to find the constants of integration when solving the differential equation.
• They ensure that the solution accurately describes the real-world behavior starting from time zero.

In our calculation, after finding the general solution of the differential equation for the initial period \( 0 \leq t < 5 \), the initial conditions are used to determine the values of constants such as \( A \) and \( B \), ensuring the solution fits the given electrical circuit's initial state.

Piecewise functions are used to define different expressions or functions for different intervals in a domain. In the context of our problem, the voltage \( E(t) \) applied to the circuit is defined as a piecewise function:

• 10 volts for \( 0 \leq t < 5 \)
• 0 volts for \( t \geq 5 \)

This means that we have two separate parts of the solution to consider because the behavior of the circuit will change at \( t=5 \) when the voltage drops to zero.
To solve these piecewise problems:

• First, solve the differential equation for the interval \( 0 \leq t < 5 \) using the given voltage.
• Then, solve the equation for \( t \geq 5 \) with the new conditions, applying continuity conditions from the initial interval to ensure a smooth transition.

Each part needs to be solved as separate initial value problems due to the changing voltage.

In the process of solving second-order differential equations, we often determine two components: the complementary solution and the particular solution.

• The **complementary solution** \( q_c(t) \) is the solution to the associated homogeneous equation (when \( f(t) = 0 \)). For example, using the characteristic equation derived from \( \frac{1}{2}q''(t) + 10q'(t) + 100q(t) = 0 \), complex roots \( \lambda = -10 \pm 10i \) lead us to a complementary solution of the form: \[ q_c(t) = e^{-10t}(A\cos(10t) + B\sin(10t)) \]
• The **particular solution** \( q_p(t) \) satisfies the entire differential equation, often including a forcing function. To find this when \( E(t) = 10 \), assume a constant solution \( q_p(t) = K \) which yields \( K = 0.1 \).

The overall solution is a combination of both:
\[ q(t) = q_c(t) + q_p(t) \]
This principle allows us to handle both homogeneous and non-homogeneous parts of the equation effectively.