Problem 57
Question
Slope Field In Exercises \(57-60\) , use a computer algebra system to graph the slope field for the differential equation and graph the solution satisfying the specified initial condition. $$ \begin{array}{l}{\frac{d y}{d x}=\frac{10}{x \sqrt{x^{2}-1}}} \\\ {y(3)=0}\end{array} $$
Step-by-Step Solution
Verified Answer
To solve the problem, plot the slope field based on the given differential equation. Then, graph the solution to the differential equation that satisfies the initial condition \(y(3) = 0\).
1Step 1: Plot the Slope Field
Use a computer algebra system to create a slope field. A slope field is a graphical representation of the differential equation, with arrows representing the slope at different points. The differential equation here is \(\frac{dy}{dx} = \frac{10}{x \sqrt{x^2-1}}\). The differential equation represents the slope, so for each point (x,y), plot an arrow with slope corresponding to the value calculated using the differential equation.
2Step 2: Graph the Solution
Now we need to find the particular solution for the differential equation that satisfies the initial condition \(y(3) = 0\). Plotting the solution curve is essentially graphing the function obtained by integrating the differential equation subject to the given initial condition.
3Step 3: Verify the Solution
Check that the function indeed satisfies differential equation and initial condition. Substitute x into the solution function, and observe that the function returns a value that verifies the initial condition \(y(3) = 0\).
Key Concepts
Slope FieldInitial ConditionGraphical RepresentationIntegration of Differential Equations
Slope Field
A slope field is a powerful visual tool that helps us understand differential equations. It represents the possible slopes of solutions to a differential equation at various points. Simply put, a slope field consists of small line segments or arrows that indicate the slope of the derivative at any given point.
For the differential equation \(\frac{dy}{dx} = \frac{10}{x \sqrt{x^2-1}}\), each line segment in the slope field shows the slope determined by plugging specific \(x\) and \(y\) values into the derivative equation.
This graphical representation is useful as it provides an overview of how solutions behave without needing the exact solution.
For the differential equation \(\frac{dy}{dx} = \frac{10}{x \sqrt{x^2-1}}\), each line segment in the slope field shows the slope determined by plugging specific \(x\) and \(y\) values into the derivative equation.
This graphical representation is useful as it provides an overview of how solutions behave without needing the exact solution.
- The slopes are drawn at various points to illustrate the direction a solution curve would take if it passed through those points.
- Slope fields give us immediate insights into the nature of solutions, such as identifying potential equilibrium points or assessing how sensitive solutions are to initial conditions.
Initial Condition
In the context of differential equations, an initial condition is an extra piece of information that allows us to find a specific solution to a differential equation.
While a differential equation might have many potential solutions, the initial condition narrows down these possibilities to one unique solution.
For example, in our exercise, the initial condition \(y(3) = 0\) is provided. This tells us that at \(x = 3\), the value of \(y\) must be 0.
While a differential equation might have many potential solutions, the initial condition narrows down these possibilities to one unique solution.
For example, in our exercise, the initial condition \(y(3) = 0\) is provided. This tells us that at \(x = 3\), the value of \(y\) must be 0.
- This initial condition helps us lock in the particular solution that goes through the point (3, 0) on the graph.
- Without an initial condition, we would find a general solution, which includes a constant of integration \(C\), and we would not be able to define \(C\) uniquely.
Graphical Representation
Graphical representation is a crucial step when dealing with differential equations. It involves plotting the solution of the differential equation that satisfies our initial condition onto the slope field we created.
This combination allows us to see how the solution curve aligns with the direction of slopes indicated in the slope field.
In the exercise, the solution curve would pass through the initial condition at \((3, 0)\) and follow the direction of arrows in the slope field.
This combination allows us to see how the solution curve aligns with the direction of slopes indicated in the slope field.
In the exercise, the solution curve would pass through the initial condition at \((3, 0)\) and follow the direction of arrows in the slope field.
- Graphical representation helps us visualize not just the solution, but also understand the dynamics of the system described by the differential equation.
- It demonstrates how the integrative process works by showing continuity and how the trajectory evolves.
Integration of Differential Equations
Integration of a differential equation is the method used to find the general solution. It's the process of reversing differentiation, which allows us to recover the original function from its derivative.
In our example, integrating \(\frac{dy}{dx} = \frac{10}{x \sqrt{x^2-1}}\) gives us a family of solutions with an arbitrary constant \(C\).
This transition is crucial in many applications, from physics to economics, where knowing the state of a system at a particular instant in time, defined by the initial condition, is essential.
In our example, integrating \(\frac{dy}{dx} = \frac{10}{x \sqrt{x^2-1}}\) gives us a family of solutions with an arbitrary constant \(C\).
- This constant is determined using the initial condition \(y(3) = 0\), helping us find a particular solution.
- The process involves using integration techniques possibly including substitution and recognizing standard integral forms.
This transition is crucial in many applications, from physics to economics, where knowing the state of a system at a particular instant in time, defined by the initial condition, is essential.
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Problem 57
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