Problem 40 - Advanced Engineering Mathematics

Question

When all the curves in a family \(G\left(x, y, c_{1}\right)=0\) intereect orthogonally, all the curves in another family \(H\left(x, y, c_{2}\right)=0\), the families are said to be orthogonal trajectories of each other. See FICURE 2.R.11. If \(d y / d x=f(x, y)\) is the differential equation of one family, then the differential equation for the orthogonal trajectores of this family is \(d y / d x=-1 / f(x, y)\). In Problems 39 and 40 , find the differential equation of the given family. Find the orthogonal trajectories of this family. Use a graphing utility to graph both families on the same set of coordinate axes. FGURE 2.I.11 Orthogonal trajectories $$ y=\frac{1}{x+c_{1}} $$

Step-by-Step Solution

Answer

The differential equations are \( \frac{dy}{dx} = -y^2 \) for family one and \( \frac{dy}{dx} = \frac{1}{y^2} \) for the orthogonal trajectories.

Step 1: Determine Differential Equation for the Given Family

The given equation is a family of curves, \( y = \frac{1}{x + c_1} \). First, differentiate this equation with respect to \(x\). Using implicit differentiation, we obtain \( \frac{dy}{dx} = -\frac{1}{(x + c_1)^2} \). However, for our purposes, we express \(c_1\) in terms of \(y\) and \(x\): \( c_1 = \frac{1}{y} - x \). To find the differential equation, differentiate \(y \cdot (x + c_1) = 1\) implicitly to obtain \( \frac{dy}{dx} = -y^2 \), which simplifies our description.

Step 2: Determine Differential Equation for Orthogonal Trajectories

The orthogonal trajectories of the family involve finding a new differential equation that is perpendicular to the original. From the original differential equation \( \frac{dy}{dx} = -y^2 \), the orthogonal trajectories will have the differential equation \( \frac{dy}{dx} = \frac{1}{y^2} \), which is the negative reciprocal of the original family’s differential equation.

Step 3: Separate Variables and Integrate

To solve the differential equations, we separate variables: Starting with the original \( \frac{dy}{dx} = -y^2 \), rearrange to \( \int y^{-2} \, dy = \int -1 \, dx \), which yields \( y^{-1} = -x + C_1 \). For the orthogonal trajectory \( \frac{dy}{dx} = \frac{1}{y^2} \), separate as \( \int y^2 \, dy = \int 1 \, dx \), yielding \( \frac{y^3}{3} = x + C_2 \).

Step 4: Solve and Find General Solutions

The integration from the separated variables provides the general solutions: For the first family, we have \( y = \frac{1}{x + C_1} \). For the orthogonal trajectories, solve \( y^3 = 3x + C_2 \) for \( y \), which simplifies to \( y = \sqrt[3]{3x + C_2} \). These represent the two families of curves.

Key Concepts

Differential Equations

In mathematics, differential equations play a crucial role in understanding how quantities change in relation to one another. A differential equation involves functions and their derivatives, describing how one variable changes with respect to another. In our case, we deal with a family of curves defined by an equation:- Given family: \[ y = \frac{1}{x + c_1} \] - The task is to find a differential equation for these curves. By differentiating with respect to \( x \), we use implicit differentiation (more on that later) to obtain: \[ \frac{dy}{dx} = -y^2 \]This equation represents the rate of change of \( y \) with respect to \( x \) in this family of curves. Once we find this differential equation, the next step is to determine the orthogonal trajectories, another set of curves that intersect our original family perpendicularly. This process involves taking the negative reciprocal of the original differential equation, which gives us:- Orthogonal trajectories' equation: \[ \frac{dy}{dx} = \frac{1}{y^2} \]

Implicit Differentiation

Implicit differentiation is a powerful technique used when dealing with equations that cannot be easily solved for one variable. It allows us to find the derivative of a function even when it is not given in the typical explicit form \( y = f(x) \).- In our exercise, we repeatedly use implicit differentiation- To find \( \frac{dy}{dx} \) for equation \( y = \frac{1}{x + c_1} \), we express \( c_1 \) in terms of \( x \) and \( y \)- This gives us the exact form to apply implicit differentiation on both sides of the equationThe equation \( y \cdot (x + c_1) = 1 \) is differentiated with respect to \( x \) using implicit differentiation, yielding:- The derivative of \( y \) with respect to \( x \) by applying the rule is: \[ \frac{dy}{dx} = -y^2 \] This step is crucial for understanding the relationship between \( x \) and \( y \) as it provides the needed differential equation to describe the family of curves.

Variable Separation

Variable separation is an essential method for solving differential equations easily. It involves rearranging the equation so that each function and its corresponding differential is isolated on one side. Subsequently, one can integrate both sides independently.- Consider the differential equation for the original family: \( \frac{dy}{dx} = -y^2 \)- Rearrange to isolate variables: \[ \int y^{-2} \, dy = \int -1 \, dx \] From this separation, you integrate both sides to find the solution:- Solving gives: \[ y = \frac{1}{x + C_1} \]- Apply the same process for orthogonal trajectories:For the orthogonal equation \( \frac{dy}{dx} = \frac{1}{y^2} \):- Rearrange and integrate: \[ \int y^2 \, dy = \int 1 \, dx \] After integration, we find the general solution:- Solve for \( y \): \[ y = \sqrt[3]{3x + C_2} \]Using variable separation helps simplify complex equations and leads to the actual solution representing the family of curves and their orthogonal trajectories.