Problem 2 - Differential Equations and Linear Algebra

Question

Determine a basis for the solution space of the given differential equation. $$y^{\prime \prime}+6 y^{\prime}+9 y=0$$

Step-by-Step Solution

Answer

The basis for the solution space of the given differential equation \(y^{\prime \prime}+6 y^{\prime}+9 y=0\) is \(\{e^{-3x}, xe^{-3x}\}\).

Step 1: Write down the given differential Equation.

The given differential equation is: \[ y^{\prime \prime} + 6y^{\prime} + 9y = 0 \]

Step 2: Find the characteristic equation.

We can replace \(y^{\prime \prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1. The characteristic equation is a quadratic equation of the form: \[ r^2 + 6r + 9 = 0 \]

Step 3: Solve the characteristic equation for its roots.

This quadratic equation can be factored as \((r + 3)^2 = 0\). Therefore, it has a repeated root: \[ r = -3 \]

Step 4: Find the linearly independent solutions of the differential equation.

Since we have a repeated root, the two linearly independent solutions of the differential equation are of the form: \[ y_1(x) = e^{-3x} \] and \[ y_2(x) = xe^{-3x} \]

Step 5: Determine the basis for the solution space of the given differential equation.

The basis for the solution space of the given differential equation is formed by the two linearly independent solutions found in step 4: \[ \{e^{-3x}, xe^{-3x}\} \] So, the basis for the solution space of the given differential equation is \(\{e^{-3x}, xe^{-3x}\}\).

Key Concepts

Characteristic Equation

In solving differential equations, the characteristic equation plays a fundamental role. It comes from converting a differential equation into an ordinary algebraic equation using substitutions. Given the second-order linear homogeneous differential equation\[y^{\prime \prime} + 6y^{\prime} + 9y = 0\],you transform the terms: replace the second derivative \(y^{\prime \prime}\) with \(r^2\), the first derivative \(y^{\prime}\) with \(r\), and the function \(y\) with 1. By doing these substitutions, you obtain the characteristic or auxiliary equation:\[r^2 + 6r + 9 = 0\].
This characteristic equation is a quadratic equation. Solving it allows us to determine the behavior of the solutions of the differential equation. In this example, it factors into \((r + 3)^2 = 0\), which reveals a repeated root.

Solution Space

The solution space of a differential equation refers to the set of all possible solutions that satisfy the given equation. For a homogeneous differential equation like\[y^{\prime \prime} + 6y^{\prime} + 9y = 0\],the solution space is a vector space. It consists of linear combinations of the basis solutions we find. To construct the solution space, we look for linearly independent solutions derived from the characteristic equation.
Given a differential equation with a repeated root from the characteristic equation, like \(r = -3\) here, we take a special approach. We start with one solution, \(y_1(x) = e^{-3x}\), and find another linearly independent solution through a modified function, \(y_2(x) = xe^{-3x}\). These provide a basis for the solution space, which in this context, is represented by the set \(\{e^{-3x}, xe^{-3x}\}\). This means any solution to the differential equation can be expressed as a combination of these two solutions.

Linearly Independent Solutions

Linearly independent solutions are crucial in understanding differential equations. Imagine you are handling a set of vectors—the principle here is similar, but with functions. Two functions are linearly independent if no scalar multiple of one function can form the other.
For a second-order homogeneous differential equation with a repeated root, such as the one we're examining, finding two linearly independent solutions ensures the complete representation of the solution space. Traditionally, repeated roots lead us to solutions of the form \(y_1(x) = e^{rx}\) and \(y_2(x) = xe^{rx}\).

• For the equation \(r = -3\), this translates to \(y_1(x) = e^{-3x}\).
• The second, linearly independent solution becomes \(y_2(x) = xe^{-3x}\) due to the repeated nature of the root.

These solutions form a basis, meaning they span the possible solutions for the differential equation, providing insights into the behavior of systems described by the equation. Understanding this relationship is key to mastering solutions in differential equations.