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Magazine Chapter 9: Problem 9
Check that \(y=A+C e^{k t}\) is a solution to the differential equation. $$\frac{d y}{d t}=k(y-A)$$
Short Answer
Expert verified
Yes, \(y = A + C e^{kt}\) is a solution to the differential equation.
Step by step solution
01
Differentiate the Given Function
To check if \(y = A + C e^{k t}\) is a solution, first find \(\frac{d y}{d t}\). We differentiate \(y = A + C e^{k t}\) with respect to \(t\), yielding \(\frac{d y}{d t} = C k e^{k t}\) because the derivative of \(C e^{k t}\) is \(C k e^{k t}\), and \(A\) is a constant, so its derivative is zero.
02
Substitute into the Differential Equation
Substitute both \(y = A + C e^{k t}\) and \(\frac{d y}{d t} = C k e^{k t}\) into the differential equation \(\frac{d y}{d t} = k(y - A)\). This gives us:\[ C k e^{k t} = k((A + C e^{k t}) - A). \]
03
Simplify the Equation
Simplify the right side of the equation:\[ k((A + C e^{k t}) - A) = k(C e^{k t}). \]Therefore, the equation becomes:\[ C k e^{k t} = k C e^{k t}. \]Both sides are equal, confirming that the original function is indeed a solution to the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solution Verification
When given a differential equation, one important step is verifying whether a proposed function is indeed a solution. For the function \( y = A + C e^{k t} \), our task is to check if it satisfies the differential equation \( \frac{d y}{d t} = k(y - A) \). Solution verification ensures the function behaves according to the rules defined by the differential equation. Here's how it's generally done:
- Differentiate the function: Begin by finding the derivative of the proposed solution with respect to the independent variable. In our case, differentiating \( y = A + C e^{k t} \) gives \( \frac{d y}{d t} = C k e^{k t} \).
- Substitute into the original equation: Replace terms in the differential equation with the expressions from differentiation and the proposed function. For our example, substitute \( y = A + C e^{k t} \) and \( \frac{d y}{d t} = C k e^{k t} \). Then simplify the equation.
- Confirm equality: Simplify both sides of the equation to check if both sides are equal. If they are, the function satisfies the differential equation.
Exponential Functions
Exponential functions, like \( e^{k t} \) in our example equation, are a cornerstone in calculus and differential equations. They help model processes that grow or decay at rates proportional to their current value, which is common in natural phenomena.
- Characteristics of exponential functions: An exponential function has the form \( y = A e^{k t} \), where \( A \) and \( k \) are constants. The base \( e \) is a mathematical constant approximately equal to 2.718. It is ubiquitous in continuous growth processes.
- Growth or decay: The sign of \( k \) determines whether the function represents growth (\( k > 0 \)) or decay (\( k < 0 \)). For example, compound interest and population growth are modeled by exponential growth functions, while radioactive decay follows an exponential decay model.
- Derivatives of exponential functions: The derivative of the exponential function \( C e^{k t} \) is \( C k e^{k t} \), showing it scales by its rate of change. This property is what makes exponential functions so useful in differential equations.
First-Order Differential Equations
First-order differential equations involve the first derivative of a function and are foundational in understanding how processes change over time. Let's explore these equations more deeply.
- Form of first-order differential equations: These equations usually appear in the form \( \frac{d y}{d t} = f(t, y) \), where \( f(t, y) \) could be a linear or nonlinear function of \( y \) and possibly \( t \). The equation \( \frac{d y}{d t} = k(y - A) \) is a simple linear first-order differential equation.
- Solving methods: First-order differential equations can often be solved using separation of variables, integrating factors, or substitution methods. In our case, recognizing the separable nature helped in solving it.
- Applications and significance: These equations are vital in fields like physics, engineering, and economics. They model diverse phenomena such as cooling, heating, population dynamics, and more.
