Chapter 20: Problem 3
Find a first-order equation satisfied by \(x=A \mathrm{e}^{-2 t}\).
Short Answer
Expert verified
The first-order differential equation is \(\frac{dx}{dt} + 2x = 0\).
Step by step solution
01
Identify the Given Function
We are provided with the function \(x = A \mathrm{e}^{-2t}\), where \(A\) is a constant, and \(x\) is a function of \(t\). Our goal is to find a first-order differential equation that this function satisfies.
02
Differentiate the Function
To find a differential equation, start by differentiating \(x\) with respect to \(t\). The derivative is: \( \frac{dx}{dt} = -2A \mathrm{e}^{-2t} \).
03
Express the Derivative in Terms of Original Function
Recognize that the original function \(x = A \mathrm{e}^{-2t} \) can be rewritten to help simplify expressions. We know:\(A \mathrm{e}^{-2t} = x\).Substitute this into the derivative: \( \frac{dx}{dt} = -2x \).
04
Write the First-Order Differential Equation
Rearrange the expression from the previous step to form a standard first-order differential equation, which in this case is:\( \frac{dx}{dt} + 2x = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a process in mathematics used to find the rate at which a quantity changes. This process, known as taking the derivative, plays a key role in understanding how functions behave. When we differentiate a function, we calculate the slope of its graph at any given point.
In our specific exercise, the function provided is: \[x = A \mathrm{e}^{-2t}\]. Differentiating this with respect to the variable \(t\), we can find how \(x\) changes as \(t\) changes.
- The derivative of \(\mathrm{e}^{-2t}\) with respect to \(t\) is \(-2 \mathrm{e}^{-2t}\) because of the chain rule.- Multiplying by the constant \(A\), the derivative becomes \(-2A \mathrm{e}^{-2t}\).Thus, the rate of change of \(x\) with respect to time \(t\) is given by \( \frac{dx}{dt} = -2A \mathrm{e}^{-2t} \). Differentiation helps us find this relationship, which is crucial for developing our differential equation.
In our specific exercise, the function provided is: \[x = A \mathrm{e}^{-2t}\]. Differentiating this with respect to the variable \(t\), we can find how \(x\) changes as \(t\) changes.
- The derivative of \(\mathrm{e}^{-2t}\) with respect to \(t\) is \(-2 \mathrm{e}^{-2t}\) because of the chain rule.- Multiplying by the constant \(A\), the derivative becomes \(-2A \mathrm{e}^{-2t}\).Thus, the rate of change of \(x\) with respect to time \(t\) is given by \( \frac{dx}{dt} = -2A \mathrm{e}^{-2t} \). Differentiation helps us find this relationship, which is crucial for developing our differential equation.
Exponential Function
An exponential function is a mathematical function of the form \(f(t) = a \cdot \mathrm{e}^{kt}\), where \(a\) and \(k\) are constants, and \(e\) is the base of natural logarithms, approximately equal to 2.71828.
Exponential functions are common in modeling real-world phenomena involving growth or decay, such as population growth, radioactive decay, and in this case, a decaying function of time. The hallmark of exponential functions is that they change proportionally to their current value.
In our exercise, we have \(x = A \mathrm{e}^{-2t}\), meaning as time \(t\) increases:
Exponential functions are common in modeling real-world phenomena involving growth or decay, such as population growth, radioactive decay, and in this case, a decaying function of time. The hallmark of exponential functions is that they change proportionally to their current value.
In our exercise, we have \(x = A \mathrm{e}^{-2t}\), meaning as time \(t\) increases:
- The value of \(x\) decreases.
- The negative exponent \(-2t\) indicates exponential decay.
Separation of Variables
Separation of variables is a technique used to solve first-order differential equations, where the variables can be isolated on opposite sides of the equation.
For the equation \(\frac{dx}{dt} + 2x = 0\), we can rearrange terms to separate the function of \(x\) from the function of \(t\).
- First, by moving \(2x\) to the other side, we get \(\frac{dx}{dt} = -2x\).- This allows us to express the relationship with differentials: \(\frac{1}{x}dx = -2 dt\).The principle of separation of variables means each side of this equation can be integrated independently:
For the equation \(\frac{dx}{dt} + 2x = 0\), we can rearrange terms to separate the function of \(x\) from the function of \(t\).
- First, by moving \(2x\) to the other side, we get \(\frac{dx}{dt} = -2x\).- This allows us to express the relationship with differentials: \(\frac{1}{x}dx = -2 dt\).The principle of separation of variables means each side of this equation can be integrated independently:
- Integrate \(\frac{1}{x} dx\) to get \(\ln|x| + C_1\).
- Integrate \(-2 dt\) to get \(-2t + C_2\).