Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Verify that \(y(t)=C e^{k t}\) is a solution to the differential equation \(\frac{d y}{d t}=k y\). (b) Verify that \(y=k e^{t}\) is not a solution to \(\frac{d y}{d t}=k y\). (c) Verify that \(y=e^{k t}+C\) is not a solution to \(\frac{d y}{d t}=k y\).

Step-by-Step Solution

Answer

The function \(y(t)=C e^{k t}\) is a solution to the differential equation, but \(y=k e^{t}\) and \(y=e^{k t}+C\) are not.

Step 1: Verify Solution 1

Differentiate \(y(t)=C e^{k t}\) using the chain rule to get \(y'(t)=k C e^{k t}\). This can be rewritten as \(y'(t)=k y(t)\), confirming that \(y(t)=C e^{k t}\) is indeed a solution to the differential equation.

Step 2: Verify Solution 2

To check the second solution, \(y=k e^{t}\), differentiate it to get \(y'=k e^{t}\). However, this is not equal to \(k y=k k e^{t}\). Thus, \(y=k e^{t}\) is not a solution to the given differential equation.

Step 3: Verify Solution 3

Lastly, differentiate \(y=e^{k t}+C\) to get \(y'=k e^{k t}\). This doesn't equal to \(k y=k (e^{k t}+C)\). So, \(y=e^{k t}+C\) is not a solution to the differential equation.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics and appear frequently in various fields including algebra, calculus, and, notably, in solving differential equations. An exponential function can be written in the form of \( y(t) = C e^{kt} \), where \( C \) and \( k \) are constants, and \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

Understanding exponential functions is critical because they describe growth or decay processes, such as radioactive decay, population growth, or compound interest. In the context of differential equations, they often represent the solution to constant coefficient linear differential equations. The ability of exponential functions to multiply themselves by a constant factor over equal intervals of time makes them incredibly useful for modeling real-world phenomena that exhibit exponential behavior.

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. It states that if you have a function \( f(g(x)) \), where both \( f \) and \( g \) are differentiable functions, then the derivative of this composite function with respect to \( x \) is given by the product of the derivative of \( f \) with respect to \( g(x) \) and the derivative of \( g \) with respect to \( x \). Mathematically, this is expressed as \( \frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \).

The chain rule is especially important when dealing with differential equations involving exponential functions. As seen in the exercise, to verify if \( y(t) = Ce^{kt} \) is a solution to the differential equation, one must differentiate \( y(t) \). Applying the chain rule correctly confirms that the derivative of \( y(t) \) with respect to \( t \) is \( kCe^{kt} \), which corresponds to the original function multiplied by the constant \( k \), thus satisfying the differential equation \( \frac{dy}{dt} = ky \).

Solutions of Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. Solutions to differential equations are functions that satisfy the equation when substituted into it. In the exercise provided, we focus on the simple yet important first-order linear differential equation \( \frac{dy}{dt} = ky \), where \( k \) is a constant. The solutions to such equations can reveal much about the behavior of dynamic systems.

When verifying potential solutions to differential equations, as in the given examples, there are a few key points to consider. First, the solution must be a function that, when differentiated, results in the right-hand side of the equation being equal to the left-hand side when substituting the function for \( y \). Second, only certain forms will satisfy the equation, and adding arbitrary constants or multiplying by constants not intrinsic to the function's exponent may result in a function that does not satisfy the differential equation. This is why in the given exercise, only \( y(t) = Ce^{kt} \) was confirmed as a valid solution, emphasizing the precision required in solving these mathematical problems.