Question

The solution of the differential equation \(\left(1+\mathrm{y}^{2}\right)+\left(\mathrm{x}-\mathrm{e}^{(\tan )-1 \mathrm{y}}\right)(\mathrm{dy} / \mathrm{dx})=0\) is: (A) \(x \cdot e^{(\tan )-1 y}=\tan ^{-1} y+k\) (B) \(x \cdot e^{(2 \tan )-1 y}=e^{-(\tan )-1 y}+k\) (C) \(2 \mathrm{x} \cdot \mathrm{e}^{(\tan )-1 \mathrm{y}}=\mathrm{e}^{(2 \tan )-1 \mathrm{y}}+\mathrm{k}\) (D) \((\mathrm{x}-2)=\mathrm{k} \cdot \mathrm{e}^{(\tan )-1 \mathrm{y}}\)

Short answer

There might be an error in the provided options or the given differential equation as none of the options match the result of Step 1.

Step-by-step solution

Write the given differential equation in terms of dy/dx

The given differential equation is: \(\left(1+y^{2}\right)+\left(x-e^{(\tan )^{-1} y}\right)\left(\frac{dy}{dx}\right)=0\) Now, isolate \(\frac{dy}{dx}\) by subtracting \(1+y^2\) from both sides and then dividing by \(\left(x-e^{(\tan )^{-1}y}\right)\): \(\frac{dy}{dx} = -\frac{1+y^2}{x-e^{(\tan )^{-1}y}}\)

Identify the potential solution by analyzing the form of dy/dx

Looking at the available options, they seem to be written in the form of \(G(x,y) = C\), where \(G(x,y)\) is a function of \(x\) and \(y\), and \(C\) is an integration constant. We will rewrite each option into the form \(\frac{dy}{dx} = F(x,y)\), then compare with our dy/dx result from Step 1.

Rewrite options in term of dy/dx

For each of the choices, calculate the derivative with respect to x: (A) \(x \cdot e^{(\tan )^{-1}y} = \tan^{-1}y + k\) \(\frac{dy}{dx} = -\frac{x}{e^{(\tan )^{-1}y}}\) (B) \(x \cdot e^{(2\tan)^{-1}y} = e^{-(\tan)^{-1}y} + k\) \(\frac{dy}{dx} = -\frac{e^{-(\tan)^{-1}y} - k}{x \cdot e^{(2\tan)^{-1}y}}\) (C) \(2x \cdot e^{(\tan)^{-1}y} = e^{(2\tan)^{-1}y} + k\) \(\frac{dy}{dx} = -\frac{e^{(2\tan)^{-1}y} - k}{2x \cdot e^{(\tan)^{-1}y}}\) (D) \((x-2) = k \cdot e^{(\tan)^{-1}y}\) \(\frac{dy}{dx} = \frac{k \cdot e^{(\tan )^{-1}y}}{(x-2)}\)

Compare the dy/dx form of the options with the result of Step 1

Comparing the dy/dx terms: Option (A): \(\frac{dy}{dx} = -\frac{x}{e^{(\tan )^{-1}y}}\) Result from Step 1: \(\frac{dy}{dx} = -\frac{1+y^2}{x-e^{(\tan )^{-1}y}}\) The form of dy/dx is different. Option (B), (C), and (D): The forms of dy/dx are also different from the result of Step 1. None of the given options match the result from Step 1. Therefore, there might be an error in the provided options or the given differential equation.

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In the context of real-world problems, differential equations are instrumental in modeling how a physical process evolves over time.

The differential equation given in our exercise is \[\left(1+y^{2}\right)+\left(x-e^{(\tan )^{-1} y}\right)\left(\frac{dy}{dx}\right)=0\] Here, the differential equation involves an unknown function y(x), its derivative dy/dx, and an inverse trigonometric function. Students encountering this type of equation should recognize that it can often be solved by isolating the derivative and integrating both sides. Assuring that the solution involves an integration step, it is crucial for students to grasp the principles of single-variable calculus, especially how to integrate functions that involve inverses of trigonometric functions.

Integration Constant

During the integration of a differential equation, an important concept to be aware of is the integration constant, often denoted by 'k' or 'C'. An integration constant represents the infinite number of solutions that are possible when integrating a function.

For example, if you were to integrate \(\frac{dy}{dx} = 3\), integrating with respect to x gives us \(y = 3x + C\), where C is the integration constant. This constant can take any real number value, which means our solution represents a family of lines with different y-intercepts.

In the case of our exercise, after determining \(\frac{dy}{dx}\), we would integrate to find y(x), and a 'k' or 'C' should emerge naturally in the solution. When comparing solutions with those offered in multiple-choice questions, each differing only by the integration constant, the form of the solution holds more significance than the constant's value, so long as it remains arbitrary.

Inverse Trigonometric Functions

Working with Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\tan^{-1}(y)\), commonly referred to as arctangent of y, are used to determine an angle when given its tangent value. In calculus and differential equations, inverse trigonometric functions appear frequently, and understanding their derivatives is essential when solving differential equations.

They also obey certain properties; for example, the derivative of \(\tan^{-1}(y)\) with respect to y is \(\frac{1}{1+y^2}\). These functions naturally complicate the process of integrating, requiring the student to use substitution or other integration techniques.

The given differential equation utilizes these functions and thus involves using their derivatives in the process. Mastery of the mechanics involving these functions facilitates a deeper understanding of their role in the solutions of differential equations.