Question

Think About It It is known that \(y=e^{k t}\) is a solution of the differential equation \(y^{\prime \prime}-16 y=0 .\) Find the values of \(k .\)

Short answer

The values of \(k\) are 4 and -4.

Step-by-step solution

Compute the derivative of \(y\)

The derivative of the function \(y=e^{kt}\) is given by \(y'=ke^{kt}\) and the second derivative is \(y^{\prime \prime} = k^2 e^{kt}\).

Substitute derivatives into the given equation

Substitute the values of \(y\) and \(y^{\prime \prime}\) in the given equation. The equation becomes \(k^2e^{kt} - 16e^{kt} = 0\). Simplifying, we have \(k^2 - 16 = 0\).

Solve for \(k\)

Solve the equation \(k^2 - 16 = 0\). The solutions are \(k = \pm 4\), because \(4^2 = 16\).

Key concepts

Exponential Functions

Exponential functions are a special type of mathematical function where the variable appears in the exponent. The standard form of an exponential function looks like this: \( y = e^{kt} \). Here, \( e \) is a mathematical constant approximately equal to 2.71828. It's known as the base of natural logarithms and is used extensively in calculus and differential equations. Exponential functions are characterized by their rapid rate of growth or decay depending on whether the exponent is positive or negative. They are extremely useful for modeling real-world phenomena like population growth, radioactive decay, and continuously compounded interest rates.
In the context of differential equations, exponential functions often serve as solutions because they maintain their form even after differentiation. This characteristic makes them extremely handy when solving ordinary differential equations (ODEs), especially those with constant coefficients.

Second Derivative

The second derivative of a function gives us insight into the curvature or concavity of the graph of the function. In simpler terms, while the first derivative tells us how fast the function is changing, the second derivative tells us how the rate of change is itself changing. For our given function \( y = e^{kt} \), the first derivative is \( y' = ke^{kt} \) and the second derivative is \( y'' = k^2e^{kt} \).The process of finding a second derivative involves different strategies depending on the initial function. When dealing with exponential functions, as in our exercise, the differentiation process leverages the function's unique property of retaining its form after differentiation. This means that each time we differentiate \( e^{kt} \), we simply multiply by the appropriate power of \( k \).
In our problem, the second derivative is crucial in forming the original differential equation, \( y'' - 16y = 0 \), which lets us solve for \( k \).

Solving Equations

Solving equations, especially differential equations, often involves several steps to find a meaningful solution. For our differential equation \( y'' - 16y = 0 \), the task is to find the values of \( k \) where the exponential function \( y = e^{kt} \) satisfies the equation. We start by substituting the derivatives we calculated into the equation. This gives us \( k^2e^{kt} - 16e^{kt} = 0 \).Notice how the exponential term \( e^{kt} \) doesn't affect the ability to solve for \( k \) because it's a common factor and never zero for any real number \( t \). We can simplify our equation by dividing both sides by \( e^{kt} \), resulting in \( k^2 - 16 = 0 \). This is now a simple algebraic equation, which can be solved using techniques for quadratic equations. As solutions, we find \( k = \pm 4 \) because \( 4^2 - 16 = 0 \).
Equation solving in mathematics, particularly in calculus, often involves transforming complex equations into simpler forms.

Mathematical Solutions

In mathematics, finding a solution means determining the values of variables that satisfy a given equation. Solutions can be values that solve algebraic equations, integrals, or in the context of differential equations, functions.For the problem \( y'' - 16y = 0 \), the solution is the set of all \( k \) values for which \( y = e^{kt} \) satisfies the equation. Our step-by-step approach leads us to the values \( k = 4 \) and \( k = -4 \). This means both \( e^{4t} \) and \( e^{-4t} \) are valid solutions to the original differential equation.
Finding solutions to differential equations often involves identifying all potential functions that satisfy the equation's conditions. Understanding the type of solution we are seeking, whether it's a specific value or a function form, is crucial to successful problem-solving in mathematics.