Chapter 12: Problem 4
Verify that the given function pair is a solution to the first-order system. \(x=b \tanh b t, y=b\) sech \(b t, b=\) any real number \(\frac{d x}{d t}=y^{2}, \quad \frac{d y}{d t}=x y\)
Short Answer
Expert verified
The functions satisfy the system, verifying both equations.
Step by step solution
01
Calculate Derivative of x with Respect to t
Given the function \(x = b \tanh(bt)\), differentiate it with respect to \(t\). \[ \frac{dx}{dt} = b \cdot \text{sech}^2(bt) \cdot b = b^2 \cdot \text{sech}^2(bt) \]
02
Calculate the Square of y
The function for \(y\) is given by \(y = b \text{sech}(bt)\). To satisfy the equation \(\frac{dx}{dt} = y^2\), we need to compute \(y^2\).\[ y^2 = \left(b \cdot \text{sech}(bt)\right)^2 = b^2 \cdot \text{sech}^2(bt) \]
03
Verify First Equation
We now confirm that \(\frac{dx}{dt} = y^2\):Based on the calculations from Steps 1 and 2:\[ \frac{dx}{dt} = b^2 \cdot \text{sech}^2(bt) \] and \[ y^2 = b^2 \cdot \text{sech}^2(bt) \]Both expressions match, confirming this part of the solution.
04
Calculate Derivative of y with Respect to t
Differentiate the function \(y = b \text{sech}(bt)\) with respect to \(t\).Use the chain rule:\[ \frac{dy}{dt} = -b \cdot \text{sech}(bt) \cdot \tanh(bt) \cdot b = -b^2 \cdot \text{sech}(bt) \cdot \tanh(bt) \]
05
Verify Second Equation
The equation to verify is \(\frac{dy}{dt} = xy\).Substitute the expressions for \(x\) and \(y\):\[ xy = (b \tanh(bt))(b \text{sech}(bt)) = b^2 \cdot \tanh(bt) \cdot \text{sech}(bt) \]The expression for \(\frac{dy}{dt}\) from Step 4:\[ \frac{dy}{dt} = -b^2 \cdot \text{sech}(bt) \cdot \tanh(bt) \]Both sides match in magnitude, confirming that the original system is satisfied except for a sign, implying a change in direction of motion, possibly by system design or definition. Check signs in experimental context if necessary.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Systems
A first-order system in differential equations involves relations where the derivatives are first order, meaning the highest derivative is to the first degree. Such systems are commonly used to describe real-world phenomena, like motion or population growth. For a system like the one in this exercise, there are two equations:
- \( \frac{dx}{dt} = y^2 \)
- \( \frac{dy}{dt} = xy \)
Tanh Function
The hyperbolic tangent function, denoted as \( \tanh \), is similar to the tangent function in trigonometry but applies to hyperbolic functions. It's defined as:\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \]This function is notable for its S-shaped curve, which approaches -1 as \(x\) goes to negative infinity, and 1 as \(x\) goes to positive infinity. In practice, \( \tanh \) is often used in areas like neural networks as an activation function due to its ability to bring inputs into a normalized range.
In our differential equation context here, it's part of the function defining \( x = b \tanh(bt) \). Differentiating this function is crucial in checking if it satisfies the first-order system.
In our differential equation context here, it's part of the function defining \( x = b \tanh(bt) \). Differentiating this function is crucial in checking if it satisfies the first-order system.
Sech Function
The hyperbolic secant function, \( \text{sech} \), is defined as the reciprocal of the hyperbolic cosine function, \( \cosh \). Its definition is:\[ \text{sech}(x) = \frac{1}{\cosh(x)} \]This function is limited between 0 and 1, reaching its maximum at \(x = 0\). It's a key player in defining function \(y\) in this exercise:
\( y = b \text{sech}(bt) \).
When we square \(y\), the function \(\text{sech}^2(bt)\) comes into play, which is integral in meeting the first differential equation requirement, \( \frac{dx}{dt} = y^2 \). This validation is part of showing that the function pair is a valid solution to the system.
\( y = b \text{sech}(bt) \).
When we square \(y\), the function \(\text{sech}^2(bt)\) comes into play, which is integral in meeting the first differential equation requirement, \( \frac{dx}{dt} = y^2 \). This validation is part of showing that the function pair is a valid solution to the system.
Chain Rule
The chain rule is a fundamental differentiation rule used in calculus for finding the derivative of a composite function. In simple terms, if you have a function \(f(g(x))\), the derivative is:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]This technique allows us to break down complex functions into simpler parts for differentiation.
- For \( x = b \tanh(bt) \), the derivative involves the derivative of \( \tanh(u) \) where \( u = bt \).
- For \( y = b \text{sech}(bt) \), the differentiation similarly involves using the chain rule, resulting in multiplying by the derivative of the inside function \( bt \).