Question

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ y^{\prime \prime \prime}+8 y=0 $$

Short answer

Question: Find the general solution of the following third-order linear homogeneous differential equation: $$ y^{\prime \prime \prime}+8 y=0 $$ Answer: The general solution of the given differential equation is: $$ y(x) = C_1e^{2x}+ C_2e^{2x\cdot \operatorname{cis}(\frac{3\pi i}{2})} + C_3e^{2x\cdot \operatorname{cis}(-\frac{3\pi i}{2})} $$

Step-by-step solution

Identify the Characteristics Equation

Given the differential equation: $$ y^{\prime \prime \prime}+8 y=0 $$ To find the general solution, we first need to determine the characteristics equation associated with this differential equation. The characteristics equation for this differential equation is written as: $$ \lambda^3+8=0 $$

Solve the Characteristics Equation

Now, we need to solve the characteristics equation for its roots: $$ \lambda^3=-8 $$ This equation can be rewritten as: $$ \lambda^3+8i^3=0 $$ Now, let's solve for λ using the formula: $$ z = r(\operatorname{cis}{(\phi)})=(-8i)^{1/3} $$ Where our z equals λ, r = 1 as only the angle changes when raised to 1/3. Now solve for the imaginary angle. $$ \phi = \operatorname{arg}(-8i) = \frac{3\pi}{2} $$ So we solve for the three roots. $$ (-8i)^{1/3} = \operatorname{cis}\left(\frac{2\pi k + \frac{3\pi}{2}}{3}\right) = 2\operatorname{cis}\left(\frac{\pi(2k + 3)}{6}\right), \quad k=0,1,2 $$ Calculating the roots: $$ \lambda_0 = 2\operatorname{cis}\left(0\right) = 2, \\ \lambda_1 = 2\operatorname{cis}\left(\frac{9\pi}{6}\right) = 2e^{\frac{3\pi i}{2}}, \\ \lambda_2 = 2\operatorname{cis}\left(\frac{15\pi}{6}\right) = 2e^{-\frac{3\pi i}{2}} $$

Write the General Solution

Now, we can express the general solution for the given differential equation using the roots of the characteristics equation: $$ y(x) = C_1e^{\lambda_0 x} + C_2e^{\lambda_1 x} + C_3e^{\lambda_2 x} $$ Substitute the roots: $$ y(x) = C_1e^{2x}+ C_2e^{2x\cdot \operatorname{cis}(\frac{3\pi i}{2})} + C_3e^{2x\cdot \operatorname{cis}(-\frac{3\pi i}{2})} $$

Solve for the Initial Conditions (If Provided)

As no initial conditions are specified in the exercise, our final solution remains the general solution: $$ y(x) = C_1e^{2x}+ C_2e^{2x\cdot \operatorname{cis}(\frac{3\pi i}{2})} + C_3e^{2x\cdot \operatorname{cis}(-\frac{3\pi i}{2})} $$

Key concepts

General Solution

In the context of differential equations, the general solution represents a family of functions that satisfy the given equation. It encompasses all possible solutions, based on arbitrary constants that are determined by initial conditions.

For example, if we are given the differential equation \(y''' + 8y = 0\), the task is to first rewrite this into a form that can be tackled using standard methods to find the general solution. This typically involves finding a polynomial, known as the characteristic equation, whose roots will form the basis of the solution.

The solution will incorporate the exponential form of these roots, and any complex roots will introduce sine and cosine components due to Euler's formula. Thus, the general solution provides a comprehensive description of all potential behaviors of the system modeled by the differential equation.

Characteristics Equation

The characteristics equation is a crucial tool in solving linear differential equations. It is derived from the original differential equation and takes the form of a polynomial equation, whose solutions tell us about the behavior of the differential equation.

For instance, the characteristics equation for \(y''' + 8y = 0\) is \(\lambda^3 + 8 = 0\). This polynomial equation results from substituting each derivative in the differential equation with a power of \(\lambda\), assuming a solution of the form \(e^{\lambda x}\). The roots of this equation, \(\lambda\), are pivotal, as they determine the form of the general solution.

Solving this polynomial gives the actual values of \(\lambda\), and whether these are real or complex affects the form of the solution, incorporating real exponential terms or trigonometric functions.

Initial Value Problem

An initial value problem, or IVP, involves finding a specific solution to a differential equation given a set of initial conditions. These initial conditions give specific values for the function and possibly its derivatives at a given point, which uniquely determine the arbitrary constants in the general solution.

In our case, we were not provided with initial conditions, so we only found the general solution. Had initial conditions been provided, we would substitute these into the general solution to solve for the specific coefficients \(C_1, C_2,\) and \(C_3\) in the solution \(y(x)\). Each condition eliminates one arbitrary constant by providing a specific constraint, thereby "pinning down" the family of solutions to one specific solution.

• A missing piece of data makes the solution general.
• Specific initial values yield a particular solution.

Complex Roots

In the process of solving differential equations, complex roots often arise from the characteristic equation. They significantly influence the form of the solution due to their association with oscillatory behavior.

For the equation \(\lambda^3 + 8 = 0\), solving for \(\lambda\) yielded complex roots. These roots indicate a solution involving complex exponentials, which through Euler's formula are translated to functions involving sine and cosine.

For instance, a root like \(\lambda = 2e^{i\frac{3\pi}{2}}\) converts into cyclical components, resulting in \(e^{ix} = \text{cos}(x) + i\text{sin}(x)\). Thus, small oscillations or rotations are embodied in part of the solution. This demonstrates why complex roots add richness to the behavior of solutions, representing phenomena like waves or alternating currents in physical systems.

• Complex numbers show up in physical applications.
• Their real components lead to exponential growth or decay.
• Their imaginary components introduce oscillations.