Question

Find the general solution. $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=0 $$

Short answer

Question: Find the general solution to the given third-order linear homogeneous differential equation: \(y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=0\) Answer: The general solution to the given differential equation is: \(y(x) = e^x(C_1 + C_2 x + C_3 x^2)\), where \(C_1, C_2, C_3\) are constants.

Step-by-step solution

Write down the auxiliary polynomial equation

To write down the auxiliary polynomial equation, replace each y^{'} term with a power of r (the root). So, we get the following equation: $$ r^3 - 3r^2 + 3r -1 = 0 $$

Find the roots of the auxiliary polynomial equation

To find the roots, let's factor the polynomial equation. In this case, we'll notice that the equation is in the form of a factorable polynomial, which is a difference of cubes. So, we can rewrite it as follows: $$ (r - 1)(r^2 - 2r + 1) = 0 $$ Now, we have two factors. Solving the first factor: $$ r - 1 = 0 \implies r = 1 $$ Solving the second factor, which turns out to be a perfect square trinomial: $$ r^2 - 2r + 1 = (r - 1)^2 = 0 \implies r = 1 $$ All roots of the auxiliary polynomial are equal and equal to 1. So, we have a triple root at r = 1.

Write down the general solution

Since we found a triple root, the general solution to the given differential equation is: $$ y(x) = e^x(C_1 + C_2 x + C_3 x^2) $$ where \(C_1, C_2, C_3\) are constants, and \(e^x\) comes from the root \(r = 1\). This is the general solution to the given third-order linear homogeneous differential equation.

Key concepts

Auxiliary Polynomial Equation

An auxiliary polynomial equation is a tool used to solve linear differential equations, especially when dealing with constant coefficients. For differential equations, you replace each derivative term with powers of a variable, like \( r \). This simplification helps to focus on solving the structure of the equation rather than its differential nature.
For example, with the differential equation \( y''' - 3y'' + 3y' - y = 0 \), the auxiliary polynomial becomes \( r^3 - 3r^2 + 3r - 1 = 0 \).
By converting differential terms into a polynomial format, we can apply algebraic methods to find solutions. The result helps us understand the nature of the solutions to the differential equation itself.
In essence, the auxiliary polynomial equation is like a bridge that connects complex differential equations with algebraic techniques, making it easier to solve.

Roots of the Polynomial

Finding the roots of a polynomial is a crucial step in solving its corresponding differential equation. Roots tell us about the structure and number of solutions. In our example, solving \( r^3 - 3r^2 + 3r - 1 = 0 \) involves factoring the polynomial to simplify it further.
Let's break it down:

• The given polynomial can be factored using the difference of cubes, resulting in \( (r - 1)(r^2 - 2r + 1) = 0 \).
• The factor \( r^2 - 2r + 1 \) is a perfect square trinomial. It simplifies to \( (r - 1)^2 = 0 \).

Thus, both factors include the root \( r = 1 \). This multiplication of factors indicates a triple root at \( r = 1 \).
So, understanding how to find roots is essential as it dictates the form of the solution, especially when roots have specific multiplicities, like in our case.

General Solution

The general solution of a differential equation is a function or set of functions that satisfies the equation. In our case, after finding that \( r = 1 \) is a triple root, the general solution for the differential equation is structured using these roots.
For a triple root \( r = 1 \), the general solution is given by: \[y(x) = e^{x}(C_1 + C_2 x + C_3 x^2)\]where \( e^{x} \) arises from the root \( r = 1 \), and \( C_1, C_2, C_3 \) are arbitrary constants that can be determined via initial conditions.
Each exponential part \( e^{xr} \) of a root represents a different function part. When roots are repeated, as in this case, we multiply by \( x \) raised to increasing powers.
These constants and the structure of the solution are key to embracing all results that satisfy the original equation. It's fascinating how fundamental algebraic roots translate into these complex yet harmonious functions.