Question

Solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{(4)}-16 y=0, \quad y(0)=2, y^{\prime}(0)=2, y^{\prime \prime}(0)=-2, y^{\prime \prime \prime}(0)=0 $$

Short answer

Short Answer: The particular solution to the given fourth-order linear homogeneous differential equation with initial conditions is: $$y(x) = \frac{5}{8}e^{2x} + \frac{3}{8}e^{-2x} + \cos(2x)$$ To graph this solution, use a graphing tool or software to plot the function within a chosen domain (e.g., \(-2 \le x \le 2\)). The graph will portray an increasing exponential function combined with a cosine function.

Step-by-step solution

Find the general solution of the differential equation

We have a fourth-order linear homogeneous differential equation: $$y^{(4)} - 16y = 0$$ To find its general solution, let's first find the characteristic equation: $$r^4 - 16 = 0$$ Solve for r: $$r^4 = 16$$ We have: $$r = \pm 2, \pm 2i$$ These roots give us the general solution: $$y(x) = C_1 e^{2x} + C_2 e^{-2x} + C_3 \cos(2x) + C_4 \sin(2x)$$

Apply the initial conditions

Use the given initial conditions to determine the constants in the general solution: First, apply \(y(0) = 2\): $$2 = C_1 e^0 + C_2 e^0 + C_3 \cos(0) + C_4 \sin(0)$$ $$2 = C_1 + C_2 + C_3$$ Next, apply \(y'(0) = 2\). First, take the first derivative of the general solution, \(y'(x)\): $$y'(x) = 2C_1 e^{2x} - 2C_2 e^{-2x} - 2C_3 \sin(2x) + 2C_4 \cos(2x)$$ Now apply the initial condition: $$2 = 2C_1 - 2C_2 + 2C_4$$ Then, apply \(y''(0) = -2\). First, take the second derivative of the general solution, \(y''(x)\): $$y''(x) = 4C_1 e^{2x} + 4C_2 e^{-2x} - 4C_3 \cos(2x) - 4C_4 \sin(2x)$$ Now apply the initial condition: $$-2 = 4C_1 + 4C_2 - 4C_3$$ Finally, apply \(y'''(0) = 0\). First, take the third derivative of the general solution, \(y'''(x)\): $$y'''(x) = 8C_1 e^{2x} - 8C_2 e^{-2x} + 8C_3 \sin(2x) - 8C_4 \cos(2x)$$ Now apply the initial condition: $$0 = 8C_1 - 8C_2 - 8C_4$$

Find the constants

Solve the system of equations obtained from the initial conditions to find the constants \(C_1\), \(C_2\), \(C_3\), and \(C_4\): $$2 = C_1 + C_2 + C_3$$ $$2 = 2C_1 - 2C_2 + 2C_4$$ $$-2 = 4C_1 + 4C_2 - 4C_3$$ $$0 = 8C_1 - 8C_2 - 8C_4$$ Manipulating the equations and solving, we find: $$C_1 = \frac{5}{8}, C_2 = \frac{3}{8}, C_3 = 1, C_4 = 0$$

Write the particular solution

Substitute the values of the constants into the general solution to find the particular solution: $$y(x) = \frac{5}{8}e^{2x} + \frac{3}{8}e^{-2x} + \cos(2x)$$

Graph the solution

To graph the particular solution, follow the function profile of: $$y(x) = \frac{5}{8}e^{2x} + \frac{3}{8}e^{-2x} + \cos(2x)$$ The graph will consist of an increasing exponential function combined with a cosine function. Define the domain for the graph (e.g., \(-2 \le x \le 2\)) and use a graphing tool or software to plot the solution.

Key concepts

Fourth-Order Differential Equations

When dealing with differential equations, a fourth-order differential equation is a type in which the highest derivative is the fourth derivative. These equations look daunting because they involve higher levels of derivatives but follow similar solving patterns as lower-order equations.
Fourth-order differential equations are more complex because they describe systems that might have interactions in a manner requiring four separate conditions to solve. These conditions help specify a particular solution for the equation, honing in on a single direction from the general solution.
Typically, when faced with these equations, you'll often derive a characteristic equation, which helps simplify the process of finding the general solution.

Characteristic Equation

The characteristic equation is a crucial tool used to solve linear homogeneous differential equations, such as our fourth-order equation here. It's like a bridge that lets you transition from the differential equation's domain to algebra, making the problem simpler.
For the fourth-order differential equation given in the problem, the characteristic equation is derived by replacing the derivatives in the differential equation with powers of a variable, often denoted as 'r'.
In this case, the characteristic equation is \(r^4 - 16 = 0\). Solving this algebraically, we find that \(r^4 = 16\), which gives root solutions in the form of \(r = \pm 2, \pm 2i\). These roots enable building the general solution using exponential and trigonometric functions that align with these solutions.

General Solution

The general solution to a differential equation is like having a full set of possible solutions that can be customized with different constants. For the fourth-order differential equation \(y^{(4)} - 16y = 0\), the roots obtained from the characteristic equation allow us to write this general solution.
The roots \(r = \pm 2, \pm 2i\) suggest that the solution will include both exponential and trigonometric components, leading to \(y(x) = C_1 e^{2x} + C_2 e^{-2x} + C_3 \cos(2x) + C_4 \sin(2x)\). Here, \(C_1, C_2, C_3,\) and \(C_4\) are constants that are determined through the application's initial conditions.
This variety ensures that we have a suitable template to describe many potential behaviors of the system defined by the differential equation, depending on those initial conditions.

Particular Solution

A particular solution is obtained once you apply the conditions specific to the problem to the general solution. It zeroes in on one precise solution that fits the prescribed conditions.
In this exercise, initial conditions \(y(0) = 2, y'(0) = 2, y''(0) = -2,\) and \(y'''(0) = 0\) are provided. Applying these, we solve four simultaneous equations to find values for \(C_1, C_2, C_3,\) and \(C_4\). Through careful calculation, these constants are determined to be \(C_1 = \frac{5}{8}, C_2 = \frac{3}{8}, C_3 = 1,\) and \(C_4 = 0\).
By inserting these constants into the general solution, we obtain the particular solution: \(y(x) = \frac{5}{8} e^{2x} + \frac{3}{8} e^{-2x} + \cos(2x)\). This solution represents the specific behavior of the system, given the initial conditions outlined in the problem.