Question

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ y^{(4)}+2 y^{\prime \prime}+y=0 $$

Short answer

Question: Find the general solution of the following fourth-order differential equation with constant coefficients: $$y^{(4)}+2y^{\prime \prime}+y = 0$$ Answer: The general solution of the given differential equation is $$y(x) = c_1\cos{x} + c_2\sin{x} + c_3x\cos{x} + c_4x\sin{x}$$, where \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are arbitrary constants.

Step-by-step solution

Write the characteristic equation

The given differential equation can be written in the form of a characteristic equation: $$ r^4 + 2r^2 + 1 = 0 $$

Solve the characteristic equation

To solve the characteristic equation, we can perform a substitution: Let \(u = r^2\), then we have $$ u^2 + 2u + 1 = 0 $$ The equation can be factored as \((u+1)^2 = 0\). Thus, \(u = -1\) is the only solution and we restore the variable \(r\) by \(r^2 = -1\), which leads to \(r = \pm i\).

Write down the general solution

With the solutions of the characteristic equation, \(r = \pm i\), we can write down the general solution of the given differential equation: $$ y(x) = c_1\cos{x} + c_2\sin{x} + c_3x\cos{x} + c_4x\sin{x} $$ where \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are arbitrary constants. Now, suppose we are given initial conditions as well. Let's consider some general initial conditions: \(y(0)=y_0, y'(0)=y'_0, y''(0)=y''_0,\) and \(y^{(3)}(0)=y^{(3)}_0.\) With these initial conditions, we will find the specific solution for the initial-value problem.

Substitute the initial conditions into the general solution

First, let's compute the first, second, and third derivatives of \(y(x)\): $$ y'(x) = -c_1\sin{x} + c_2\cos{x} - c_3x\sin{x} + c_3\cos{x} + c_4x\cos{x} + c_4\sin{x} $$ $$ y''(x) = -2c_1\cos{x} - 2c_2\sin{x} + 2c_4\cos{x} - 2c_3\sin{x} $$ $$ y^{(3)}(x) = 2c_1\sin{x} - 2c_2\cos{x} + 2c_3\cos{x} + 2c_4\sin{x} $$ Now, we evaluate the derivatives at \(x = 0\): $$ y(0) = c_1 = y_0 \\ y'(0) = c_2 = y'_0 \\ y''(0) = -2c_1 + 2c_4 = y''_0 \\ y^{(3)}(0) = 2c_2 + 2c_3 = y^{(3)}_0 $$ Solving these equations for \(c_1, c_2, c_3, c_4\), we find: $$ c_1 = y_0 \\ c_2 = y'_0 \\ c_3 = \frac{y^{(3)}_0 - 2y'_0}{2} \\ c_4 = \frac{y''_0 + 2y_0}{2} $$

Present the specific solution for the initial-value problem

The specific solution for the initial-value problem with initial conditions \(y(0)=y_0, y'(0)=y'_0, y''(0)=y''_0,\) and \(y^{(3)}(0)=y^{(3)}_0\) is: $$ y(x) = y_0\cos{x} + y'_0\sin{x} + \frac{y^{(3)}_0 - 2y'_0}{2}x\cos{x}+ \frac{y''_0 + 2y_0}{2}x\sin{x} $$

Key concepts

Characteristic Equation

Understanding the characteristic equation is crucial when dealing with linear differential equations. In our exercise, the differential equation was given as

$$y^{(4)}+2y^{\text{''}}+y=0$$

To solve this, we first need to form the characteristic equation. This equation is derived by substituting

$$y=e^{rx}$$

into the differential equation and simplifying it. The characteristic equation for our problem is

$$r^4+2r^2+1=0$$

By solving this, we find the roots of the equation, which indicate the solution to the differential equation. In many cases, we may find real or complex roots. Here, after substitution with

$$u=r^2$$

we determined the roots to be complex,

$$r=\text{±}i$$

indicating a solution involving trigonometric functions due to the imaginary roots. It's important to understand that finding and analyzing the characteristic equation is a stepping stone to obtaining the general solution of a differential equation.

Initial Value Problem

When additional information is provided at the outset, such as the value of the function and its derivatives at a specific point, we have an initial value problem (IVP). The step-by-step solution provided delineates how to incorporate initial conditions into the general solution.

For the IVP, we are usually given the function's value and its derivatives at

$$x = 0$$

These conditions are utilized to determine the arbitrary constants present in the general solution. In other words, they anchor the solution to a particular scenario. You can think of the initial conditions as the 'starting point' for the behavior of our solution. With our exercise, by substituting these initial values into the computed derivatives and solving, we refined the general solution to a specific solution that fits the given initial conditions. This tailored solution can predict the behavior of the function given any future or past values of

$$x$$

The initial value problem, therefore, is a fundamental concept when a real-world scenario is modeled, as it personalizes the equation to fit the specifics of the situation at hand.

General Solution of Differential Equation

The general solution of a differential equation includes all possible solutions to the equation. It encompasses various relationships between the variables involved while satisfying the equation itself. In the differential equation provided, the general solution derived from the characteristic equation's complex roots is

$$y(x) = c_1\cos{x} + c_2\sin{x} + c_3x\cos{x} + c_4x\sin{x}$$

with

$$c_1, c_2, c_3, \text{and } c_4$$

being arbitrary constants that will be determined by the initial conditions, if such are given. These constants give the flexibility to the solution to encompass an infinite number of particular solutions, each suitable for different initial conditions or different scenarios. Remember, the general solution provides the complete picture of the differential equation's behavior and is crucial for understanding the system before applying any specific conditions.