Ordinary Differential Equations

We use \(q\) to denote the charge (in colombs), \(i\) to denote the current (in amperes). In a single-loop or series circuit, the sum of the voltage drops across an inductor, resistor, and capacitor is equal to the impressed voltage \(E\text{.}\)

	Inductance	Resistance	Capacitance
Units	henries \(h\)	Ohms \(\Omega\)	farads \(f\)
Electrical law	Faraday’s law \(V=L \dfrac{\mathrm{d}i}{\mathrm{d}t}\)	Ohm’s law \(V=Ri\)	Capacitance law \(V=\dfrac{1}{C}q\)



A circuit with an electrical source, an inductor, a resistor, and a capacitor in a loop back to the electrical source.

🔗

These electrical components connected in different ways lead to a variety of different possible differential equations:

Example 1.4.3.

Write a differential equation for the \(LR\) circuit for the current \(i\) when \(L=2 \text{h}\text{,}\) \(R=5 \Omega\text{,}\) \(i(0)=3 \text{amp}\text{,}\) and the impressed voltage is \(E(t)=t \text{volt}\text{.}\)

🔗

🔗

Example 1.4.4.

The \(RC\) circuit with resistance \(R=1 \Omega\) and capacitance \(C=\dfrac{1}{2} \text{f}\) when connected to an AC power source \(E(t)=\sin(5t)\) has the following differential equation for charge \(q\text{:}\)

\begin{equation*} q'+2q=\sin(5t). \end{equation*}

It can be shown that a general solution of this differential equation is \(q(t)=c_1 e^{-2t}+\dfrac{2}{29}\sin(5t)-\dfrac{5}{29}\cos(5t)\text{.}\) Plot the solution for various choices of the parameter \(C\text{.}\)

🔗

🔗