Problem 23 - Differential Equations and Linear Algebra

Question

Determine all values of the constant \(r\) such that the given function solves the given differential equation. $$y(x)=e^{r x}, \quad y^{\prime \prime}+6 y^{\prime}+9 y=0$$.

Step-by-Step Solution

Answer

The value of the constant \(r\) that makes the given function \(y(x) = e^{rx}\) a solution to the differential equation \(y'' + 6y' + 9y = 0\) is \(r = -3\).

Step 1: Find the first and second derivatives of \(y(x)\)

First, we need to find the first and second derivatives of the given function with respect to \(x\). Given function: \(y(x) = e^{rx}\) First derivative: \(y'(x) = \frac{d}{dx} (e^{rx}) = re^{rx}\) Second derivative: \(y''(x) = \frac{d^2}{dx^2} (e^{rx}) = r^2e^{rx}\) Now, we have the first and second derivatives: \(y'(x) = re^{rx}\) and \(y''(x) = r^2e^{rx}\).

Step 2: Substitute the derivatives and the function into the differential equation

Next, substitute the function \(y(x)\), its first derivative \(y'(x)\), and its second derivative \(y''(x)\) into the given differential equation. Differential equation: \(y''(x) + 6y'(x) + 9y(x) = 0\) Substituting, we get: \((r^2e^{rx}) + 6(re^{rx}) + 9(e^{rx}) = 0\)

Step 3: Solve for the constant \(r\)

Now we need to solve for \(r\) in the resulting equation. First, notice that \(e^{rx}\) is a common factor in all three terms. Divide the entire equation by \(e^{rx}\), which is nonzero for all values of \(x\), to simplify the equation: \(r^2 + 6r + 9 = 0\) At this point, we can recognize this equation as a quadratic equation in the form \(Ar^2 + Br + C = 0\), where \(A = 1\), \(B = 6\), and \(C = 9\). To solve this equation for \(r\), we can either factor it or use the quadratic formula. In this case, factoring is straightforward: \((r + 3)^2 = 0\) This equation implies that \(r + 3 = 0\), which means \(r = -3\). So, the value of the constant \(r\) that makes the given function a solution to the differential equation is \(r = -3\).

Key Concepts

Quadratic Equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown. In the exercise, we dealt with the quadratic equation \( r^2 + 6r + 9 = 0 \). Solving this equation helps in finding the values of \(r\) that ensure our function, \(y(x) = e^{rx}\), solves the differential equation given.

Quadratic equations can be solved using several methods:

• **Factoring**: This method involves expressing the quadratic equation as a product of binomials. If it can be factored easily as in \((r + 3)^2 = 0\), we directly find \(r = -3\).
• **Quadratic Formula**: If the equation is difficult to factor, use the formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• **Completing the Square**: Rewriting the equation in the form \((r + m)^2 = d\), can also be helpful.

Utilizing these methods will help solve quadratic equations efficiently, crucial for tasks involving differential equations as shown.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent, like \(y(x) = e^{rx}\). These functions grow or decay at rates proportional to their values. The constant \(e\) is approximately equal to 2.71828, known as Euler's number.

Exponential functions have unique mathematical properties:

• **Derivative**: The derivative of an exponential function \(e^{rx}\) is \(re^{rx}\), which means their rate of change is proportional to the function itself.
• **Base of Natural Logarithms**: The exponential function with base \(e\) has the natural logarithm \(\ln(x)\) as its inverse.

In differential equations, exponential functions such as \(y(x) = e^{rx}\) frequently appear as solutions to linear homogeneous equations due to their derivative properties.

Derivatives

Derivatives represent the rate at which a function changes. Given a function \(y(x) = e^{rx}\), finding its derivatives is vital in differential equations as demonstrated in the exercise.

Key points about derivatives:

• **First Derivative**: \(y'(x) = re^{rx}\) shows how fast \(y(x)\) changes with respect to \(x\).
• **Second Derivative**: \(y''(x) = r^2e^{rx}\) indicates how the rate of change itself changes, giving us insight into the concavity of \(y(x)\).

When working with differential equations, we substitute these derivatives back into the equation to check if our function, such as \(y(x) = e^{rx}\), satisfies the conditions given by the equation.

Homogeneous Equations

Homogeneous differential equations like \(y'' + 6y' + 9y = 0\) have solutions where terms are proportionate. They're characterized by having the form \(Ay'' + By' + Cy = 0\). In these equations:

Key Characteristics:

• Coefficients \(A\), \(B\), and \(C\) are real numbers or constants setting the structure of the equation.
• Solutions often involve exponential functions since their derivatives replicate their form.

In solving homogeneous equations, substitute the derivatives of the exponential function and simplify.

The goal is to identify specific values like \(r\) that make the equation valid. This identifies solutions that adhere to the relationship defined by the characteristic equation, highlighting the pattern or behavior of many physical systems described by differential equations.