Question

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

Short answer

The general solution to the 4th-order linear homogeneous differential equation is $$ y(t) = C_1 e^{t} + C_2 e^{-t} + C_3 + C_4 e^{(\frac{-t}{2})}\cos(\frac{\sqrt{3}}{2}t) + C_5 e^{(\frac{-t}{2})}\sin(\frac{\sqrt{3}}{2}t) $$

Step-by-step solution

Find the general solution to the 4th-order differential equation

First, let us find the general solution to the given 4th-order linear homogeneous differential equation: $$ y^{(4)}-y^{\prime \prime \prime}+y^{\prime}-y=0 $$ To do this, let us start by assuming a solution of the form: $$ y = e^{rt} $$ where r is a constant. Taking the first derivative, second derivative, third derivative, and fourth derivative, and plugging these back into the equation, we get: $$ (e^{rt})(r^4 - r^3 + r -1)=0 $$ We need to find the roots of this characteristic equation. We can rewrite the equation as: $$ r^4 - r^3 + r -1=0 $$ This is a quartic polynomial, and its roots can be quite complicated, but we do notice that r = 1 is a root: $$ 1 - 1 + 1 -1=0 $$ Hence, using synthetic division, we can find the depressed cubic with r=1 as root: $$ (r-1)(r^3 -1) =0 $$ By factoring the cubic further, we find two more roots, r = -1 and r = 0: $$ (r-1)(r + 1)(r^2 + r + 1) = 0 $$ Therefore, we have the general solution: $$ y(t) = C_1 e^{t} + C_2 e^{-t} + C_3 + C_4 e^{(\frac{-t}{2})}\cos(\frac{\sqrt{3}}{2}t) + C_5 e^{(\frac{-t}{2})}\sin(\frac{\sqrt{3}}{2}t) $$ where \(C_1, C_2, C_3, C_4\) and \(C_5\) are constants.

Apply initial conditions (if any) and solve the initial value problem

The problem does not specify any initial conditions. Therefore, the answer to this exercise is the general solution derived in Step 1: $$ y(t) = C_1 e^{t} + C_2 e^{-t} + C_3 + C_4 e^{(\frac{-t}{2})}\cos(\frac{\sqrt{3}}{2}t) + C_5 e^{(\frac{-t}{2})}\sin(\frac{\sqrt{3}}{2}t) $$

Key concepts

General Solution

In differential equations, a general solution is a form of the solution that encompasses all possible solutions of the equation. It includes arbitrary constants that can be altered based on specific requirements, such as initial conditions. This type of solution forms the foundation upon which specific solutions are built.

For example, the differential equation given in this exercise is \[y^{(4)} - y^{'''} + y^{'} - y = 0\]. Solving this involves finding solutions for each component of the equation depending on various assumption models, like assuming a basic form of solution such as \(y = e^{rt}\).

• The point of solving for a general solution is to make sure it operates for any scenarios or initial parameters that might be adjusted later.
• In this case, because it's a homogeneous differential equation, the general solution is composed of all possible specific solutions added together.

This means you could plug any set of initial conditions into this framework to solve any specific instance of the equation.

Initial Value Problem

An initial value problem (IVP) in mathematics is a problem where you find a solution to a differential equation that satisfies particular conditions at the start, or 'initial' values. This is particularly useful in real-world situations where certain conditions are consistent at a starting point.

When working with an IVP, your goal is to use these initial conditions to find the specific constants within a general solution, turning it into a specific solution that fits the scenario perfectly.

• In this exercise, no initial conditions were specified, so the problem does not extend beyond finding the general solution.
• However, if conditions like \(y(0) = 1\) or \(y'(0) = 2\) were known, they would be used to solve for \(C_1, C_2, C_3, C_4, \) and \(C_5\).

This part of solving a differential equation helps align the mathematical model with real-world observations, making it extremely practical.

Characteristic Equation

The characteristic equation is a critical tool when working with linear differential equations. It's derived from assuming that the solution to a differential equation has an exponential form. The characteristic equation gives you a polynomial whose roots are essential to forming the general solution.

In this exercise, the characteristic equation is \(r^4 - r^3 + r - 1 = 0\).

• Simplifying this equation helps in finding the roots, and thus the types of terms to expect in the general solution, like exponential or trigonometric functions.
• Finding the roots involves techniques like factoring or using synthetic division—as seen with the root \(r=1\), reducing the polynomial to a cubic one and so on.

By analyzing these roots, we conclude important features about the solution, such as stability or oscillatory behavior.

Quartic Polynomial

A quartic polynomial is a polynomial of degree four. In the context of differential equations, it arises when you equate a characteristic polynomial with zero to identify possible root values that will form part of the solution.

Dealing with quartic polynomials can be complex due to their higher degree, but they are manageable with the right techniques such as factorization, and sometimes numerical methods might be necessary.

• Here, the quartic polynomial is \(r^4 - r^3 + r - 1 = 0\).
• Finding a root like \(r = 1\) allows for the use of synthetic division to simplify quartic to simpler cubic or quadratic equations.

Solving these kinds of polynomials often leads to a mix of polynomial factor roots and perhaps some complex or repeated roots, all of which contribute to the form of the general solution. Handling such polynomials requires not only algebra skills but also a good grasp of estimating and dealing with complex numbers.