Problem 21 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given higher order differential equation. $$y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y=0$$

Step-by-Step Solution

Answer

The general solution is \( y = (C_1 + C_2x + C_3x^2)e^{-x} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is a linear homogeneous differential equation of third order with constant coefficients. It has the form \( y''' + 3y'' + 3y' + y = 0 \).

Step 2: Write the Characteristic Equation

For a linear homogeneous differential equation with constant coefficients, we find the characteristic equation by replacing \( y^{(n)} \) with \( r^n \). Thus, the characteristic equation for \( y''' + 3y'' + 3y' + y = 0 \) is \( r^3 + 3r^2 + 3r + 1 = 0 \).

Step 3: Find the Roots of the Characteristic Equation

We solve the characteristic equation \( r^3 + 3r^2 + 3r + 1 = 0 \). By using a method such as synthetic division or trial and error with possible roots (for example, \( r = -1 \)), we find that \( r = -1 \) is a root. Dividing \( r^3 + 3r^2 + 3r + 1 \) by \( r + 1 \) gives us \( (r+1)^3 = 0 \). Thus, \( r = -1 \) is a triple root.

Step 4: Write the General Solution

Since \( r = -1 \) is a root with multiplicity 3, the general solution is formed by \( y = (C_1 + C_2x + C_3x^2)e^{-x} \), following the formula for multiple roots with multiplicity \( k \). Here, \( C_1, C_2, \) and \( C_3 \) are arbitrary constants.

Key Concepts

Linear Homogeneous Differential Equations

A linear homogeneous differential equation is a special type of differential equation. These equations include derivatives of a function but no additional terms without the function or its derivatives. A key feature of these equations is that the terms can be added up to zero, as seen in the form \[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \ ... + a_1 y^{ ext{'}} + a_0 y = 0 \] where each term is scaled by a constant coefficient.The term "linear" indicates that each term of the equation is either a derivative of the unknown function or a constant multiplier of those derivatives, and "homogeneous" signifies that the sum of those terms is zero.Such equations can be of any order depending on the highest derivative involved. For example, the differential equation in the original exercise is of third order, meaning the highest derivative is the third derivative, \(y'''\). This classification is crucial since it helps determine the approach for finding solutions.

Characteristic Equation

To solve a linear homogeneous differential equation with constant coefficients, we use the characteristic equation. This equation is derived by replacing derivatives with powers of \(r\) in a hypothetical polynomial equation.Given a differential equation such as \[ y''' + 3y'' + 3y' + y = 0 \],we assume a solution of the form \(y = e^{rx}\) and substitute it back into the differential equation.This substitution makes the assumption that the derivatives of \(y\) will convert into powers of \(r\). The result of this process is the characteristic polynomial,\[ r^3 + 3r^2 + 3r + 1 = 0 \],which needs to be solved to find roots \(r\). These roots will help us construct the general solution for the differential equation.

Roots of Characteristic Equation

The roots of the characteristic equation are crucial for forming the general solution. They represent the exponential factors in the differential equation's solution.For the characteristic equation\[ r^3 + 3r^2 + 3r + 1 = 0, \]we seek values for \(r\) that satisfy the equation. Methods such as synthetic division or trial and error can be used to find these roots. Here, finding that \(r = -1\) is a root leads us to use division that reveals a repeated root, resulting in \[ (r + 1)^3 = 0. \]The root \(r = -1\) is a triple root, indicating that it is repeated three times. These roots are pivotal because they determine how the terms in the general solution are constructed, including the handling of repeated roots.

General Solution of Differential Equation

The general solution of a linear homogeneous differential equation is built directly from the roots of its characteristic equation.When roots are distinct, each root \(r_i\) contributes a term \(C_i e^{r_i x}\) to the solution. However, if a root has a multiplicity greater than one, the solution incorporates polynomial terms due to the repetition.For the problem, with a characteristic equation having the root \(r = -1\) of multiplicity 3, the general solution is\[ y = (C_1 + C_2 x + C_3 x^2) e^{-x}, \]where \(C_1, C_2,\) and \(C_3\) are constants determined by initial conditions (if given) or remain arbitrary otherwise.Each part of this form handles the multiplicity by including increasing powers of \(x\) coupled with \(e^{-x}\), reflecting each presence of the repeated root.