Problem 24 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ y^{\prime \prime}-6 y^{\prime}+9 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = (C_1 + C_2 t) e^{3t} \).

Step 1: Characteristic Equation

To find the general solution, start by writing the characteristic equation associated with the differential equation. For the given equation \( y'' - 6y' + 9y = 0 \), the characteristic equation is obtained by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. This gives us:\[ r^2 - 6r + 9 = 0 \]

Step 2: Solve Characteristic Equation

Next, solve the characteristic equation \( r^2 - 6r + 9 = 0 \). This is a quadratic equation, which can be factored as:\[ (r - 3)^2 = 0 \]Therefore, the roots are \( r = 3 \). Since this root is repeated, it occurs with multiplicity 2.

Step 3: Write General Solution

The general solution to a second-order linear differential equation with constant coefficients, when there is a repeated root \( r \), is given by:\[ y(t) = (C_1 + C_2 t) e^{rt} \]Since our repeated root is \( r = 3 \), the general solution for this differential equation becomes:\[ y(t) = (C_1 + C_2 t) e^{3t} \]

Key Concepts

Characteristic Equation in Linear Differential Equations

In the realm of linear differential equations, the characteristic equation plays a crucial role in finding solutions. When handling a homogeneous linear differential equation with constant coefficients, such as \( y'' - 6y' + 9y = 0 \), we derive the characteristic equation by substituting derivatives with their corresponding powers of \( r \). This means replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1.

For the equation at hand, this substitution leads to the characteristic equation \( r^2 - 6r + 9 = 0 \). Solving this equation provides the roots that dictate the form of the solution to the original differential equation.

Here are key points to remember when deriving the characteristic equation:

• Identify the order of the differential equation (second-order in our example).
• Substitute correctly: \( y'' \to r^2 \), \( y' \to r \), and \( y \to 1 \).
• Solve the resulting polynomial equation for \( r \).

Understanding the General Solution

Once the roots of the characteristic equation have been found, the next step involves constructing the general solution of the differential equation. The nature of the roots—whether they are distinct, repeated, or complex—determines the form of this general solution.

For instance, in the equation \( y'' - 6y' + 9y = 0 \), the roots found were repeated, specifically \( r = 3 \). This means the solution takes a specific form that accommodates for the repeated nature of the roots.

The general solution for a differential equation when there is a repeated root \( r \) is:

• \( y(t) = (C_1 + C_2 t) e^{rt} \)

This formula emerges because the repeated root generates a dependency between the terms of the solution. Understanding this helps students predict solutions' behavior more effectively. When applying this to our equation, the general solution becomes \( y(t) = (C_1 + C_2 t) e^{3t} \).

• \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Dealing with Repeated Roots

Repeated roots arise when the characteristic equation's discriminant equals zero, indicating a single repeated solution. In our example, the characteristic equation derived from \( y'' - 6y' + 9y = 0 \) factored to \((r - 3)^2 = 0\), revealing that \( r = 3 \) is a repeated root.

The appearance of repeated roots means that a lone exponential term is insufficient for capturing the complete family of solutions. To address this, the solution must introduce a polynomial of terms with increasing powers of \( t \) (starting with \( t^0 \) which is 1) multiplying the exponential term. This is why the general solution includes \( C_1 + C_2 t \), ensuring all possible solutions are covered by this expression.

Key points regarding repeated roots:

• Recognize that repeated roots result in a solution involving both \( e^{rt} \) and \( t \cdot e^{rt} \).
• These adjustments ensure the general solution spans all possible behaviors when the differential equation is subject to various initial conditions.