Question

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

The family of curves is given by \(\frac{1}{2}(y^2 + x^2) + xy = C\). Their orthogonal trajectories are given by \(\frac{1}{2}(x^2 - y^2) - xy = C\).

Step-by-step solution

Rewrite the Differential Equation

Given: \((x + y) dy + (x - y) dx = 0\)Rewrite it in the form \(M(x, y) + N(x, y)dx = 0\):\(M(x, y) = x + y, N(x, y) = x - y\)Therefore, the equation is:\((x + y) dy = -(x - y) dx\) or \(\frac{dy}{dx} = \frac{-(x - y)}{x + y}\)

Separate Variables

Separate the variables to solve the differential equation:\(\frac{dy}{dx} = \frac{-(x - y)}{x + y}\)Rewrite as:\(dy = -\frac{x - y}{x + y} dx\)Separate variables:\((y+x) dy = -(x - y) dx\)

Integrate Both Sides

Integrate both sides to find the family of curves:\[\int (y + x) dy = -\int (x - y) dx\]\(\frac{1}{2} y^2 + xy = -\frac{1}{2} x^2 + xy + C\)Combine terms to get:\(\frac{1}{2} y^2 + xy + \frac{1}{2} x^2 = C\)Thus, the family of curves is given by: \(\frac{1}{2} (y^2 + x^2) + xy = C\)

Find the Slope of the Orthogonal Trajectories

Orthogonal trajectories have slopes that are negative reciprocals of the original family’s slopes. Original slope is \(\frac{dy}{dx} = \frac{-(x - y)}{x + y}\). Therefore, the slope of orthogonal trajectories is \(\frac{dx}{dy} = \frac{x + y}{-(x - y)}\).

Separate Variables for Orthogonal Trajectories

Rewrite \(\frac{dx}{dy} = \frac{x + y}{-(x - y)}\)and do the variable separation:\((x - y) dx = -(x + y) dy\)

Integrate Both Sides for Orthogonal Trajectories

Integrate both sides:\[\int (x - y) dx = -\int (x + y) dy\]\(\frac{1}{2} x^2 - xy = -\frac{1}{2} y^2 - xy + C\)Combine terms to get:\(\frac{1}{2} x^2 - xy + \frac{1}{2} y^2 = C\)Thus, the orthogonal trajectories are given by: \(\frac{1}{2} (x^2 - y^2) - xy = C\)

Key concepts

headline of the respective core concept

Orthogonal trajectories are curves that intersect a given family of curves at right angles (90 degrees). To find these orthogonal trajectories, we need to find a new family of curves whose slopes are the negative reciprocals of the original family’s slopes. This means if the slope of the original family of curves is given by \(\frac{dy}{dx} = f(x, y)\), then the slope of the orthogonal trajectories will be \(\frac{dx}{dy} = -\frac{1}{f(x, y)}\).In the given problem, the original slope is \(\frac{dy}{dx} = \frac{-(x - y)}{x + y}\). Thus, the slope of the orthogonal trajectories can be written as \(\frac{dx}{dy} = \frac{x + y}{-(x - y)}\). - These orthogonal trajectories help to visualize how different curves cross each other perpendicularly. - They also have practical applications in physics, engineering, and even gaming.

headline of the respective core concept

Variable separation is a method used to solve ordinary differential equations by separating the variables on different sides of the equation. This makes it easier to integrate and find the solution. In our given problem, the equation \(\frac{dy}{dx} = \frac{-(x - y)}{x + y}\) was first rewritten as \(dy = -\frac{x - y}{x + y} dx\).Next, we separated the variables:\(\frac{(y + x)}{dy} = - (x - y) dx\) - This allows us to integrate both sides separately. - It's crucial to be careful with algebraic manipulations during variables separation to avoid mistakes.

headline of the respective core concept

Integration is a fundamental concept in calculus used to find the area under a curve or to compute antiderivatives. Once we have separated the variables, the next step is to integrate both sides to solve the differential equations.In our problem, after separating the variables, we integrate both sides:\( \int (y + x) dy = - \int (x - y) dx \).The result of this integration is:\( \frac{1}{2} y^2 + xy = - \frac{1}{2} x^2 + xy + C \)Combining terms, we get the general solution for the family of curves:\( \frac{1}{2} ( y^2 + x^2 ) + xy = C \).This integration gives us the family of curves expressed in a single equation. Similarly, for the orthogonal trajectories, we integrate again after separating variables to find their specific equations. - This step-by-step integration is vital to transforming a differential equation into a more usable form, simplifying the understanding of multiple interconnected curves.