Problem 15 - Differential Equations and Linear Algebra

Question

Determine the general solution to the given differential equation. $$\left(D^{2}-2\right) y=0$$

Step-by-Step Solution

Answer

The general solution of the given differential equation \(\left(D^{2}-2\right) y=0\) is: \[y(x) = C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x}\]

Step 1: Identify the operator and write down the differential equation explicitly

We are given a differential equation in the form of an operator, which can be written more explicitly as: \[y'' - 2y = 0\] where \(D^2\) represents the second derivative operator, and \(y''\) denotes the second derivative of the function \(y(x)\) with respect to \(x\).

Step 2: Write down the characteristic equation

Next, we will convert this differential equation into a characteristic equation. We do this by replacing the second derivative term \(y''\) with \(r^2\) and the original function \(y(x)\) with 1. So, the characteristic equation becomes: \[r^2 - 2 = 0\]

Step 3: Solve the characteristic equation

Now, we need to solve the characteristic equation to find the roots. These roots will help us write down the general solution of the differential equation. To solve the quadratic equation \(r^2 - 2 = 0\), we can use the quadratic formula or factoring. In this case, we can easily find the roots: \[r^2 = 2 \Rightarrow r = \pm \sqrt{2}\]

Step 4: Write down the general solution

Since we have two real and distinct roots, the general solution can be written in the following form: \[y(x) = C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x}\] where \(C_1\) and \(C_2\) are arbitrary constants. So, the general solution of the given differential equation \(\left(D^{2}-2\right) y=0\) is: \[y(x) = C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x}\]

Key Concepts

Characteristic Equation

In the study of differential equations, the characteristic equation plays a crucial role in finding the solution for linear homogeneous differential equations. To form this equation, we treat each derivative as a power of a variable, normally denoted by 'r'. For the given differential equation \(y'' - 2y = 0\), the corresponding characteristic equation is formed by replacing \(y''\) with \(r^2\) and the function \(y\) with 1. This yields \(r^2 - 2 = 0\).

Solving the characteristic equation is similar to solving any quadratic equation and typically involves factoring or using the quadratic formula. The roots of the characteristic equation, in this case \(r = \pm\sqrt{2}\), play a critical role in determining the form of the general solution to the differential equation. If these roots are real and distinct, as they are here, the solution will involve exponential functions with the roots in the exponents.

Second Derivative

The term second derivative, notated as \(y''\) or \(\frac{d^2y}{dx^2}\), signifies the rate at which the first derivative of a function changes. It provides us with information about the curvature and concavity of the function's graph. In our differential equation \(y'' - 2y = 0\), the presence of the second derivative indicates that we are dealing with a second-order differential equation. Second-order differential equations often describe motion in physics, such as simple harmonic motion.

In the solution process, the second derivative is central to forming the characteristic equation and thus leading us toward the general solution. It's important to understand that the second derivative not only tells us about the steepness of the graph (as the first derivative does) but also how that steepness is changing over time or space.

General Solution

The general solution to a differential equation encompasses all possible solutions and includes arbitrary constants, usually denoted by \(C_1\), \(C_2\), etc. These constants can later be determined if initial conditions are provided. For the equation at hand, with second derivative terms and constants, the general solution is \(y(x) = C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x}\).

This form of the solution is reflective of the fact that our characteristic equation yielded real and distinct roots. Each term in the solution corresponds to one root of the characteristic equation, and these terms are linearly independent. The exponential functions arise because they are the only functions whose derivatives are proportional to the function itself, making them a natural fit for linear homogeneous differential equations such as this one. Understanding the structure of the general solution is essential for further study of differential equations, including applications to real-world scenarios.