Second-order linear differential equations with constant coefficients have a variety of applications in science and engineering. Some of the key applications include:
- Electrical Circuits
- Simple Harmonic Motion
- Bending of Beams
Electrical Circuits (L-C-R Circuits)
Consider an electric circuit containing a resistance \( R \), an inductance \( L \), and a capacitance \( C \). If \( i \) is the current in the circuit at any time \( t \), then:
- The potential drop across the resistance is given by: \[ E_R = iR \quad \text{(Ohm’s Law)}. \]
- The potential drop across the inductance is given by: \[ E_L = L \frac{di}{dt}. \]
- The potential drop across the capacitance is given by: \[ E_C = \frac{q}{C}, \] where \( q \) is the charge in the circuit.
Kirchhoff’s Voltage Law states that "In an electric circuit, the sum of voltage drops (potential drops) is equal to the supplied voltage (e.m.f)." Mathematically, this can be expressed as: \[ E_R + E_L + E_C = \text{Supplied Voltage (e.m.f)}. \]
Differential Equation for L-C-R Circuits
Using Kirchhoff’s Voltage Law, the equation for an L-C-R circuit can be written as: \[ L \frac{di}{dt} + Ri + \frac{q}{C} = E(t), \] where \( E(t) \) is the supplied voltage as a function of time.
Since \( i = \frac{dq}{dt} \), the equation can be rewritten in terms of charge \( q \): \[ L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = E(t). \] This is a second-order linear differential equation with constant coefficients.
The solution to this differential equation consists of two parts:
- Complementary Function (C.F.): The solution to the homogeneous equation (when \( E(t) = 0 \)).
- Particular Integral (P.I.): A specific solution to the non-homogeneous equation (when \( E(t) \neq 0 \)).
The complete solution is given by: \[ q(t) = \text{C.F.} + \text{P.I.} \]
Example 1. In an \( L-C-R \) circuit, the charge \( q \) on a plate of the condenser is given by: \[ L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = E \sin pt. \] The circuit is tuned to resonance so that \( p^2 = \frac{1}{LC} \). If initially the current \( i \) and the charge \( q \) are zero, show that for small values of \( \frac{R}{L} \), the current in the circuit at time \( t \) is given by: \[ i = \frac{Et}{2L} \sin pt. \]
Solution: The differential equation governing the \( L-C-R \) circuit is: \[ \left( L D^2 + R D + \frac{1}{C} \right) q = E \sin pt, \] where \( D = \frac{d}{dt} \).
The auxiliary equation is: \[ L m^2 + R m + \frac{1}{C} = 0. \] Solving for \( m \): \[ m = \frac{1}{2L} \left( -R \pm \sqrt{R^2 - \frac{4L}{C}} \right). \] Substituting \( p^2 = \frac{1}{LC} \), we get: \[ m = -\frac{R}{2L} \pm \sqrt{\frac{R^2}{4L^2} - p^2}. \] For small \( \frac{R}{L} \), the term \( \frac{R^2}{4L^2} \) is negligible, and the roots are approximately: \[ m \approx -\frac{R}{2L} \pm ip. \]
The complementary function (C.F.) is: \[ q_{\text{CF}} = e^{-\frac{Rt}{2L}} \left( c_1 \cos pt + c_2 \sin pt \right). \] For small \( \frac{R}{L} \), we can approximate \( e^{-\frac{Rt}{2L}} \approx 1 - \frac{Rt}{2L} \).
The particular integral (P.I.) is: \[ q_{\text{PI}} = \frac{E}{R} \frac{1}{D} \sin pt = -\frac{E}{Rp} \cos pt. \]
The complete solution is: \[ q = \left( 1 - \frac{Rt}{2L} \right) \left( c_1 \cos pt + c_2 \sin pt \right) - \frac{E}{Rp} \cos pt. \quad \text{(1)} \]
The current \( i = \frac{dq}{dt} \) is: \[ i = \left( 1 - \frac{Rt}{2L} \right) \left( -c_1 p \sin pt + c_2 p \cos pt \right) - \frac{R}{2L} \left( c_1 \cos pt + c_2 \sin pt \right) + \frac{E}{R} \sin pt. \quad \text{(2)} \]
Initially, \( q = 0 \) and \( i = 0 \) at \( t = 0 \). Substituting these into (1) and (2), we get: \[ c_1 = \frac{E}{Rp}, \quad c_2 = \frac{E}{2L p^2}. \]
Substituting \( c_1 \) and \( c_2 \) into (2), we get: \[ i = \left( 1 - \frac{Rt}{2L} \right) \left( -\frac{E}{Rp} \sin pt + \frac{E}{2L p^2} \cos pt \right) p - \frac{R}{2L} \left( \frac{E}{Rp} \cos pt + \frac{E}{2L p^2} \sin pt \right) + \frac{E}{R} \sin pt. \] Simplifying: \[ i = \frac{Et}{2L} \sin pt - \frac{E}{4Rp} \left( \frac{R}{L} \right)^2 \cos pt - \frac{E}{4R p^2} \left( \frac{R}{L} \right)^2 \sin pt. \] For small \( \frac{R}{L} \), the terms involving \( \left( \frac{R}{L} \right)^2 \) are negligible, and we get: \[ i = \frac{Et}{2L} \sin pt. \]
Model Problems
Solve the following differential equations:
-
Exercise: Find the current \(i\) in the \(L - C - R\) circuit assuming zero initial current and charge. Given: \(L = 20 \, \text{henries}\), \(C = 0.01 \, \text{farads}\), \(R = 80 \, \text{ohms}\), and \(E = 100 \, \text{volts}\).
Answer: The current \(i(t)\) is given by: \[ i(t) = \frac{E}{R} \left( 1 - e^{-\frac{R}{2L}t} \cos(\omega t) \right), \] where \(\omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}\). -
Exercise: An inductor of 2 henries, resistor of 16 ohms, and capacitor of 0.02 \(mF\) are connected in series with a battery of e.m.f. \(E = 100 \sin 3t\). At \(t = 0\), the charge of the capacitor and current in the circuit are zero. Find the charge and current at \(t > 0\).
Answer: The charge \(q(t)\) and current \(i(t)\) are given by: \[ q(t) = A e^{-\alpha t} \cos(\omega t + \phi) + \frac{E_0}{\sqrt{(R^2 + (\omega L - \frac{1}{\omega C})^2}} \sin(3t - \theta), \] \[ i(t) = \frac{dq}{dt}, \] where \(\alpha = \frac{R}{2L}\), \(\omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}\), and \(\theta = \tan^{-1}\left(\frac{\omega L - \frac{1}{\omega C}}{R}\right)\). -
Exercise: A condenser of capacity \(C\) is discharged through an inductance \(L\) and resistance \(R\) in series. The charge \(q\) at a time \(t\) satisfies the equation:
\[
L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0.
\]
Given \(L = 0.25 \, H\), \(R = 250 \, \Omega\), \(C = 2 \times 10^{-6} \, F\), and \(\frac{dq}{dt} = 0\) at \(t = 0\), obtain the value of \(q\) in terms of \(t\).
Answer: The charge \(q(t)\) is given by: \[ q(t) = q_0 e^{-\frac{R}{2L}t} \cos(\omega t), \] where \(\omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}\) and \(q_0\) is the initial charge on the capacitor.