Problem 10 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given second-order differential equation. $$3 y^{\prime \prime}+y=0$$

Step-by-Step Solution

Answer

The general solution is \( y(x) = c_1 \cos\left(\frac{x}{\sqrt{3}}\right) + c_2 \sin\left(\frac{x}{\sqrt{3}}\right) \).

Step 1: Understand the Differential Equation

The differential equation given is a second-order homogeneous linear differential equation of the form \[ 3 y'' + y = 0 \]. This indicates that we're looking for a function \( y(x) \) that satisfies this relationship.

Step 2: Write the Characteristic Equation

To solve the differential equation, write the characteristic equation associated with it by assuming a solution of the form \( y = e^{rx} \). Substituting this into the equation gives us the characteristic equation\[ 3r^2 + 1 = 0 \].

Step 3: Solve the Characteristic Equation for Roots

The characteristic equation is \( 3r^2 + 1 = 0 \). Solving for \( r \), we get:\[ r^2 = -\frac{1}{3} \]\[ r = \pm i\frac{1}{\sqrt{3}} \].These roots are complex, which will guide the form of the solution.

Step 4: Write the General Solution

For complex roots of the form \( r = \, \pm i\frac{1}{\sqrt{3}} \), the general solution is given by\[ y(x) = c_1 \cos\left(\frac{x}{\sqrt{3}}\right) + c_2 \sin\left(\frac{x}{\sqrt{3}}\right) \],where \( c_1 \) and \( c_2 \) are arbitrary constants.

Key Concepts

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a type of differential equation where each term is a multiple of the unknown function or its derivatives. In our exercise, we have the equation \(3 y^{\prime \prime} + y = 0\). This is called homogeneous because there is no standalone constant term or function of \(x\) on the right-hand side of the equation. Everything equals zero. This means the "output" or "response" of the equation is solely driven by the linear terms of the equation itself. Homogeneous linear equations often show up in natural processes, where things aren't being added or removed from the system, resulting in zero forcing functions (right-hand side equals zero).

Characteristic Equation

To solve a homogeneous linear differential equation, you often use its characteristic equation. This step transforms the problem into a more manageable algebraic one. It starts by hypothesizing a solution in the form \(y = e^{rx}\). Deriving the differential form and substituting it back into the differential equation leads to the characteristic equation. For the exercise, substituting gives us \[3r^2 + 1 = 0\]. Solving this equation involves finding the values of \(r\) that satisfy it, which are the roots of the equation. By solving these roots, we find crucial information about the behavior of the system modeled by the original differential equation.

Complex Roots

When solving the characteristic equation, sometimes you end up with complex roots. This happens when the discriminant (the part under the square root in the quadratic formula) is negative. Complex roots usually appear in conjugate pairs, such as \(r = \pm i\frac{1}{\sqrt{3}}\) from our example. Complex roots are significant because they suggest oscillatory solutions, commonly seen in wave and vibration problems. When dealing with them, you use Euler’s formula to transform the exponential solution into a trigonometric one, involving sines and cosines. These functions embody the oscillation patterns inherent in solutions with imaginary components.

General Solution

The general solution of a differential equation represents the full set of all possible solutions, encapsulating every way the system could behave under various initial conditions. For second-order equations with complex roots, the general solution takes a trigonometric form. In our example, the roots \(r = \pm i\frac{1}{\sqrt{3}}\) yield a general solution of \(y(x) = c_1 \cos\left(\frac{x}{\sqrt{3}}\right) + c_2 \sin\left(\frac{x}{\sqrt{3}}\right)\). Here, \(c_1\) and \(c_2\) are constants that are determined by the initial conditions of the system or context in which the equation is applied. This trigonometric format is very insightful as it suggests the system's responses are periodic, making these solutions especially useful in fields like engineering and physics.