Question

A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-\frac{d y}{d x}\left(e^{x}+e^{-x}\right)+1=0\) is given by (1) \(y+e^{-x}=C\) (2) \(y-e^{-x}=C\) (3) \(y+e^{x}=C\) (4) \(y-e^{2 x}=C\)

Short answer

The correct solutions are (2) and (3) from the given options.

Step-by-step solution

Simplify the Differential Equation

Consider the given differential equation: \[\left(\frac{dy}{dx}\right)^{2} - \frac{dy}{dx} \left(e^{x} + e^{-x}\right) + 1 = 0\] Let's denote \( \frac{dy}{dx} \) as \( p \). The equation becomes \[ p^2 - p(e^{x} + e^{-x}) + 1 = 0 \]

Solve the Quadratic Equation

Solve for \( p \) in the quadratic equation: \[ p^2 - p(e^{x} + e^{-x}) + 1 = 0 \] Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -(e^{x} + e^{-x}) \), and \( c = 1 \), we obtain \[ p = \frac{e^{x} + e^{-x} \pm \sqrt{(e^{x} + e^{-x})^2 - 4}}{2} \]

Simplify the Expression under the Square Root

Simplify the term under the square root: \[ (e^{x} + e^{-x})^2 - 4 \] This is equivalent to \[ e^{2x} + 2 + e^{-2x} - 4 = e^{2x} + e^{-2x} - 2 \]Thus, \( p = \frac{(e^{x} + e^{-x}) \pm \sqrt{e^{2x} + e^{-2x} - 2}}{2} \)

Observe Simplifications

Notice that \[ e^{2x} + e^{-2x} - 2 = (e^{x} - e^{-x})^2 \] Therefore, the equation becomes \[ p = \frac{(e^{x} + e^{-x}) \pm |e^{x} - e^{-x}|}{2} \]

Consider the Cases for Solution

Case 1: When the positive sign is taken, \[ p = \frac{(e^{x} + e^{-x}) + (e^{x} - e^{-x})}{2} = e^{x} \] Case 2: When the negative sign is taken, \[ p = \frac{(e^{x} + e^{-x}) - (e^{x} - e^{-x})}{2} = e^{-x} \]

Integrate to Find y

Case 1: Integrating \( p = e^{x} \), we get \[ y = \frac{dy}{dx} = e^{x} \Rightarrow y = e^{x} + C \] Case 2: Integrating \( p = e^{-x} \), we get \[ y = \frac{dy}{dx} = e^{-x} \Rightarrow y = e^{-x} + C \]

Match with Given Solutions

From the integral results, the possible solutions are \[ y - e^{x} = C \] and \[ y - e^{-x} = C \] Hence, the correct options are (2) and (3).

Key concepts

IIT JEE Preparation

Preparing for the IIT JEE exam can be challenging, but understanding key concepts in calculus and differential equations is crucial for success. The IIT JEE tests problem-solving abilities and comprehension of mathematical theories. Focus on mastering each concept by breaking it down into simple, understandable parts. Practice is key—you should solve as many problems as possible to become comfortable with different types of questions. Remember, analytical thinking and a methodical approach will help you excel. Also, make use of past papers and mock tests to familiarize yourself with the exam format.
Stay curious and keep revisiting difficult concepts until they click. Participation in study groups can also help you gain different perspectives on problem-solving techniques. Never hesitate to ask for help when needed. Consistency and persistence will make you well-prepared for the exam.

Calculus

Calculus is the mathematical study of continuous change. It is divided mainly into two branches: differential calculus and integral calculus. Differential calculus focuses on rates of change, such as derivatives, while integral calculus deals with the accumulation of quantities, such as integrals.
Understanding the foundational concepts of calculus is essential. For differential equations, we often use derivatives to describe how a function changes. Mastering these concepts requires understanding the notations and the rules for differentiation and integration. Visualizing functions and their changes using graphs can be extremely helpful.
In the context of solving differential equations, recognizing patterns and applying the right techniques are important. Ensure you understand the basic principles of differentiation and integration. Practice exercises that apply these concepts in various scenarios to build a strong foundation.
Take learning step by step. Start with simple problems and gradually progress to more complex ones. This approach will build your confidence and proficiency in calculus.

Integration Techniques

Integration is a fundamental concept in calculus used to find the accumulation of quantities. To solve real-world problems, different integration techniques are used, such as substitution, integration by parts, and partial fractions.
In the exercise provided, integrating both sides of the equation after recognizing the patterns leads to finding the solution. Substitution is a common method where you substitute a part of the integrand to simplify the integral. Integration by parts is used when the integrand is a product of two functions. Partial fraction decomposition is used to break down complex rational functions into simpler fractions.
The key is to recognize the appropriate technique for the problem at hand. In the provided solution, recognizing that the integration can be simplified by substituting the expressions with easier terms was crucial. Practice each technique individually and then combine them in more complex problems.
Remember to always check your solutions by differentiating the result to see if you get the original function. This helps confirm the correctness of your integration. Continuous practice and exploration of various problems will enhance your integration skills.