Question

The general solution of $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=-\omega^{2} x $$ is \(x=A \mathrm{e}^{j \omega t}+B \mathrm{e}^{-\mathrm{j} \omega t}\), where \(\mathrm{j}^{2}=-1\). Verify that this is indeed a solution. What is the particular solution satisfying \(x(0)=0\), \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0)=1 ?\) Express the general solution and the particular solution in terms of trigonometrical functions.

Short answer

Question: Verify the given general solution is a solution to the given differential equation, find the particular solution satisfying the initial conditions, and express the solutions in terms of trigonometric functions: Differential equation: \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} = -\omega^{2}x\) General solution: \(x(t) = A \mathrm{e}^{j \omega t} + B \mathrm{e}^{-\mathrm{j} \omega t}\) Initial conditions: \(x(0) = 0\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0) = 1\) Answer: The general solution is verified. The particular solution is given by \(x_p(t) = \frac{1}{2\mathrm{j}\omega} \mathrm{e}^{j\omega t} - \frac{1}{2\mathrm{j}\omega} \mathrm{e}^{-j\omega t}\), and both the general and particular solutions have been expressed in terms of trigonometric functions.

Step-by-step solution

Verify the General Solution

Before verifying that the given general solution is a solution to the given differential equation, let's first differentiate the solution twice, so we have expressions for \(\frac{\mathrm{d} x}{\mathrm{~d} t}\) and \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\): \( \frac{\mathrm{d} x}{\mathrm{~d} t} = A \mathrm{j}\omega \mathrm{e}^{j \omega t} - B \mathrm{j}\omega \mathrm{e}^{-\mathrm{j} \omega t}\) \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} = -A \omega^{2} \mathrm{e}^{j \omega t} - B \omega^{2} \mathrm{e}^{-\mathrm{j} \omega t}\) Now, plug the expression for \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\) into the given differential equation, and observe whether it holds true: \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} = -\omega^{2}x\) \( -A \omega^{2} \mathrm{e}^{j \omega t} - B \omega^{2} \mathrm{e}^{-\mathrm{j} \omega t} = -\omega^{2}(A \mathrm{e}^{j \omega t} + B \mathrm{e}^{-\mathrm{j} \omega t})\) Since both sides of the equation match, this verifies that the given general solution is indeed a solution to the given differential equation.

Find the Particular Solution

To find the particular solution, we must satisfy the given initial conditions: \(x(0) = 0\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0) = 1\) First, let's plug in \(t=0\) to both the given general solution equations for \(x(t)\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}(t)\): \(x(0) = A \mathrm{e}^{j \omega (0)} + B \mathrm{e}^{-\mathrm{j} \omega (0)} = A + B\) \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0) = A \mathrm{j}\omega \mathrm{e}^{j \omega (0)} - B \mathrm{j}\omega \mathrm{e}^{-\mathrm{j} \omega (0)} = \mathrm{j}\omega A - \mathrm{j}\omega B\) Using the initial conditions, we get: \(A + B = 0\) \(\mathrm{j}\omega A - \mathrm{j}\omega B = 1\) Solving for A and B, we get: \(A = -B\) \(\mathrm{j}\omega A - \mathrm{j}\omega (-A) = 1\) \(A = \frac{1}{2\mathrm{j}\omega}\) and \(B = -\frac{1}{2\mathrm{j}\omega}\) Thus, the particular solution is: \(x_p(t) = \frac{1}{2\mathrm{j}\omega} \mathrm{e}^{j\omega t} - \frac{1}{2\mathrm{j}\omega} \mathrm{e}^{-j\omega t}\)

Express in Trigonometric Functions

To express the general solution and particular solution in terms of trigonometric functions, we use the Euler's formula: \(\mathrm{e}^{jx} = \cos{x} + \mathrm{j}\sin{x}\) So, the general solution becomes: \(x(t) = A(\cos{\omega t}+\mathrm{j}\sin{\omega t}) + B(\cos{-\omega t}-\mathrm{j}\sin{-\omega t})\) The particular solution becomes: \(x_p(t) = \frac{1}{2\mathrm{j}\omega} (\cos{\omega t}+\mathrm{j}\sin{\omega t}) - \frac{1}{2\mathrm{j}\omega} (\cos{-\omega t}-\mathrm{j}\sin{-\omega t})\) Now, we have expressed the general solution and the particular solution in terms of trigonometric functions.

Key concepts

Second-Order Differential Equation

A second-order differential equation is a relation involving an unknown function, its derivatives, and a variable. In the context of Newtonian mechanics, second-order differential equations frequently describe motion under the influence of forces. The general second-order linear differential equation takes the form:
\[\begin{equation}a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)\text{,}\tag{1}\text{,}\tag{1}\text{,}\text{,}\text{,}\text{,}\text{,}\text{,}c d^2y\( in the exercise corresponds to acceleration, while the coefficient \)\omega^2$ represents the square of angular frequency in oscillatory systems.
These equations are particularly important in physics and engineering, describing systems ranging from simple harmonic oscillators to electric circuits and planetary motion.

Complex Numbers

Complex numbers extend the familiar real number system to include solutions to equations like \(x^2+1=0\). A complex number is expressed as \(a+bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property that \(i^2=-1\). In our exercise, we use \(j\) instead of \(i\) to represent the imaginary unit, respecting the convention in engineering fields.
Complex numbers are crucial in solving differential equations with oscillatory solutions, as they elegantly capture both growth and oscillation through the real and imaginary parts.

Euler's Formula

Euler's formula, a cornerstone in the field of complex analysis, provides a profound link between exponential functions and trigonometric functions. It states that for any real number \(x\):
\(e^{ix} = \cos(x) + i\sin(x)\)
This formula allows for converting complex exponentials into trigonometric expressions, which simplifies calculations and provides intuition for interpreting solutions to differential equations involving complex numbers. The exercise utilizes Euler's formula to express the solution in terms of sines and cosines, which are more intuitive for most physical problems.

Initial Value Problem

An initial value problem is a differential equation coupled with a set of conditions specifying the state of the system at a single point, usually at \(t=0\). For a second-order differential equation, these conditions are typically the initial displacement and initial velocity, corresponding to \(x(0)\) and \(dx/dt|_{t=0}\), respectively. Solving an initial value problem helps find a particular solution that fits a specific scenario. By using \(x(0)=0\) and \(dx/dt(0)=1\), we find constants \(A\) and \(B\) in our exercise, determining the model's behavior at the outset.

Trigonometric Functions

Trigonometric functions, including sine and cosine, describe the relationships between the angles and sides of triangles. More broadly, they are used to model periodic phenomena across sciences: from simple pendulum movements to the behavior of electromagnetic waves. By expressing the solutions to differential equations in terms of trigonometric functions, we often gain clearer insights into the periodic nature and characteristics of the solution, such as amplitude and frequency. In the given exercise, Euler's formula aids in rewriting the complex exponential terms of the solution into trigonometric functions, establishing a form familiar to students and easier to interpret in most practical applications.