Question

Find the general solution of the given differential equation. $$ y^{\mathrm{vi}}-y^{\prime \prime}=0 $$

Short answer

#Answer# The general solution for the given differential equation is: $$ y(x) = C_1 + C_2 e^{x} + C_3 e^{-x} + C_4 e^{ix} + C_5 e^{-ix} $$ where \(C_1, C_2, C_3, C_4,\) and \(C_5\) are arbitrary constants.

Step-by-step solution

Rewrite the given equation as a characteristic equation

Since the given equation contains the sixth and second derivatives of y, we can rewrite it as a characteristic equation by assuming a solution of the form \(y = e^{rx}\), where r is a constant. The characteristic equation for the given differential equation is: $$ r^6 - r^2 = 0 $$

Factor the characteristic equation

We can factor the characteristic equation as follows: $$ r^2(r^4 - 1) = 0 $$

Find the roots of the characteristic equation

The characteristic equation has roots when \(r^2 = 0\) or when \((r^4 - 1) = 0\). The roots are: $$ r = 0, \pm 1, \pm i $$

Use the roots to find the general solution

Now that we have the roots of the characteristic equation, we can use them to write the general solution for the given differential equation. The general solution is given by: $$ y(x) = C_1 e^{0x} + C_2 e^{x} + C_3 e^{-x} + C_4 e^{ix} + C_5 e^{-ix} $$ where \(C_1, C_2, C_3, C_4,\) and \(C_5\) are arbitrary constants. Since \(e^{0x} = 1\), we can simplify the general solution as: $$ y(x) = C_1 + C_2 e^{x} + C_3 e^{-x} + C_4 e^{ix} + C_5 e^{-ix} $$

Key concepts

Characteristic Equation

Whenever you encounter a differential equation with constant coefficients, a powerful method to find its solution is through the concept of the characteristic equation. This equation emerges from substituting a trial solution, typically of the form \(y = e^{rx}\) where \(r\) is the constant to be determined, into the differential equation. The purpose of this substitution is to reduce the differential equation to a polynomial whose roots will guide us in finding the general solution. For example, if you have the differential equation \(y^{\mathrm{vi}}-y^{\prime \prime}=0\), you'd get a characteristic equation of \(r^6 - r^2 = 0\) after applying the trial solution. This polynomial will then be factored, as seen in the next step.
As an educational guide, understanding how to formulate and solve characteristic equations is essential for tackling various kinds of differential equations, providing a pathway from complex calculus to more algebraic manipulation.

General Solution

The general solution of a differential equation encapsulates all possible solutions to the equation. It usually involves a combination of functions and incorporates arbitrary constants which represent the flexibility you have in choosing specific solutions. These constants are typically determined by initial conditions or boundary values. For instance, in the solution to our differential equation, the general solution is expressed as \(y(x) = C_1 + C_2 e^{x} + C_3 e^{-x} + C_4 e^{ix} + C_5 e^{-ix}\), where each term corresponds to a root of the characteristic equation and each \(C\) is an arbitrary constant.
It's important to note that the purpose of the general solution is to cover all possibilities. When solving practical problems, you'll often have additional information that will help you find the unique solution that fits the scenario, which involves determining the values of these constants. For students, mastering the composition of the general solution is key to progressing in differential equations.

Arbitrary Constants

In the realm of differential equations, arbitrary constants play a crucial role. These constants appear in the general solution of a differential equation as representatives of the infinite number of solutions that the equation can have. Each arbitrary constant corresponds to the integration of the differential equation and reflects the initial conditions that can be applied to the problem at hand. In our example, the constants \(C_1, C_2, C_3, C_4,\) and \(C_5\) are introduced when determining the general solution.
For students delving into differential equations, grasping the concept of arbitrary constants is fundamental. It's like knowing that you have multiple keys on a keyring—each key (constant) can unlock a different door (solution) under various conditions. These constants are the final piece of the puzzle when you have more information in the form of initial conditions or boundary values, allowing you to fit the general solution to a specific situation.

Roots of Equation

The roots of the characteristic equation are the cornerstones of determining the general solution to a differential equation. By identifying the roots, you essentially unravel the structure of the solution. They are the values of \(r\) that make the characteristic equation true—in other words, equal to zero. Depending on the type of roots (real or complex), the corresponding terms in the general solution will differ, such as exponential functions for real roots or sinusoidal functions for complex roots.
In practice, once the characteristic equation is crafted, as done in step 2 of our example where we factored \(r^6 - r^2 = 0\) into \(r^2(r^4 - 1) = 0\), finding the roots becomes an exercise in algebra. For the mentioned equation, the roots are discovered to be \(r = 0, \pm 1, \pm i\), which then translates directly into the exponential and trigonometric functions in the general solution. Students need to realize that the nature of these roots governs the type of solutions to expect, and so understanding how to find and interpret them is an essential skill in differential equations.