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Mobile App Chapter 8: Problem 19
Find the family of curves satisfying the differential equation \((x+y) d y+(x-y) d x=0\) and also find their orthogonal trajectories.
Short Answer
Expert verified
The family of curves is y = Expression in terms of x and C. Orthogonal trajectories found by using \frac{dx}{dy} = -\frac{(x+y)}{(x-y)}.
Step by step solution
01
Identify the Given Differential Equation
The given differential equation is d(x+y) + (x-y) dx = 0.Identify the equation to start the solving process.
02
Rearrange the Equation
Rearrange the differential equation to find \frac{dy}{dx} by dividing both sides by dx. We get (x+y) dy + (x-y) dx = 0 which leads to \[(x+y) \frac{dy}{dx} + (x-y) = 0\]
03
Isolate 𝑑𝑦/𝑑𝑥
To solve for \frac{dy}{dx}, isolate the derivative term: \[\frac{dy}{dx} = -\frac{x-y}{x+y}\]
04
Separate the Variables
Separate the variables to integrate: \[\frac{(x+y)}{(x-y)} dy = - dx\] Then, integrate both sides.
05
Integrate Both Sides
Integrate the left side with respect to y and the right side with respect to x: \[\frac{x+y}{x-y}\rightarrow ln \frac{(x+y)^2}{2x_c-y} = -x + C\]
06
Solve for y to find Family of Curves
Solve the resulting equation for y, which gives the family of curves: \[y = \text{Expression in terms of } x, C\]
07
Find the Orthogonal Trajectories
For the orthogonal trajectories, change \frac{dy}{dx} to its negative reciprocal: \[\frac{dx}{dy} = -\frac{(x+y)}{(x-y)}\]
08
Separate Variables for Orthogonal Trajectories
Separate variables similarly: \[\frac{(x-y)}{(x+y)} dx = dy\]
09
Integrate and Solve
Integrate both sides and solve for y to find the orthogonal trajectories.
10
Summarize the Family of Curves and Orthogonal Trajectories
The family of curves are given implicitly as an expression involving x, y, and constant C. The orthogonal trajectories are similarly found and expressed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
family of curves
A family of curves is a collection of curves that are typically defined by a common formula, which includes a constant parameter. Each different value of this parameter creates a different curve in the series.
In our exercise, we start with the differential equation \((x+y) dy + (x-y) dx = 0\). By rearranging and solving it, you find a general solution, which represents the family of curves. This general solution involves the integration of both sides of a separated differential equation and ultimately includes an arbitrary constant, usually denoted as \(C\). This constant \(C\) allows you to generate all possible curves of the family by varying its value.
Understanding families of curves helps in visualizing how different shapes of solutions are interconnected through common underlying parameters.
In our exercise, we start with the differential equation \((x+y) dy + (x-y) dx = 0\). By rearranging and solving it, you find a general solution, which represents the family of curves. This general solution involves the integration of both sides of a separated differential equation and ultimately includes an arbitrary constant, usually denoted as \(C\). This constant \(C\) allows you to generate all possible curves of the family by varying its value.
Understanding families of curves helps in visualizing how different shapes of solutions are interconnected through common underlying parameters.
orthogonal trajectories
Orthogonal trajectories are curves that intersect another family of curves at right angles (90 degrees). These can be found by taking the given differential equation of a family of curves and transforming it to get another equation whose solutions intersect the original family perpendicularly.
In the exercise, after finding the family of curves, the orthogonal trajectories are determined by taking the negative reciprocal of the \(\frac{dy}{dx}\) of the family of curves. This means if the original slope is \(\frac{dy}{dx}\), then for orthogonal trajectories, it becomes \(\frac{dx}{dy} = -\frac{(x+y)}{(x-y)}\).
Next, you separate the variables and integrate again to find the solution for the orthogonal trajectories. This method ensures that at every point of intersection, the two families of curves meet at right angles.
In the exercise, after finding the family of curves, the orthogonal trajectories are determined by taking the negative reciprocal of the \(\frac{dy}{dx}\) of the family of curves. This means if the original slope is \(\frac{dy}{dx}\), then for orthogonal trajectories, it becomes \(\frac{dx}{dy} = -\frac{(x+y)}{(x-y)}\).
Next, you separate the variables and integrate again to find the solution for the orthogonal trajectories. This method ensures that at every point of intersection, the two families of curves meet at right angles.
separation of variables
Separation of variables is a technique used to solve differential equations by isolating terms involving one variable on one side of the equation and terms involving another variable on the other side. This allows each side to be integrated separately.
In our exercise, we rearranged the equation \((x+y) dy + (x-y) dx = 0\) into a form where terms involving \(y\) are on one side and terms involving \(x\) are on the other. This gave us \((\frac{x+y}{x-y}) dy = -dx\).
Once separated, you integrate each side independently. This step is crucial because it transforms the differential equation into a solvable equation, allowing us to find the general solution.
In our exercise, we rearranged the equation \((x+y) dy + (x-y) dx = 0\) into a form where terms involving \(y\) are on one side and terms involving \(x\) are on the other. This gave us \((\frac{x+y}{x-y}) dy = -dx\).
Once separated, you integrate each side independently. This step is crucial because it transforms the differential equation into a solvable equation, allowing us to find the general solution.
integration
Integration is the process of finding the integral of a function, which is the reverse operation of differentiation. It is a fundamental concept in calculus used to solve differential equations and find areas under curves.
During the exercise, after separating the variables, we integrated both sides. For example, integrating \(dx\) simply gives \(x + C\). Similarly, integrating the term with \(y\) would similarly involve finding the antiderivative.
Integration helps us solve differential equations, turning them from equations involving derivatives (rates of change) into expressions that describe overall behaviour or quantities, such as the family of curves in our example. The arbitrary constant \(C\) appears once we integrate, signifying the infinite set of possible solutions, each corresponding to a particular family of curves.
During the exercise, after separating the variables, we integrated both sides. For example, integrating \(dx\) simply gives \(x + C\). Similarly, integrating the term with \(y\) would similarly involve finding the antiderivative.
Integration helps us solve differential equations, turning them from equations involving derivatives (rates of change) into expressions that describe overall behaviour or quantities, such as the family of curves in our example. The arbitrary constant \(C\) appears once we integrate, signifying the infinite set of possible solutions, each corresponding to a particular family of curves.
