Problem 102 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the differential equation \(\left(1+y^{2}\right)\left(x-e^{2 \tan ^{-1} y}\right) \frac{d y}{d x}=0\), is \(\quad\) (A) \((x-2)=k e^{-\tan ^{-1} y}\) (B) \(2 x e^{\tan ^{-1}} y=e^{2 \tan ^{-1}}+k\) (C) \(x e^{\operatorname{lan}^{-1} y}=\tan ^{-1} y+k\) (D) \(x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k\)

Step-by-Step Solution

Answer

Option D is correct: \(x e^{2 \tan^{-1} y} = e^{\tan^{-1} y} + k\).

Step 1: Understand and Simplify the Equation

The given differential equation is \((1+y^2)(x-e^{2 \tan^{-1}y}) \frac{dy}{dx} = 0\). This implies that either \((1+y^2)=0\) or \((x-e^{2 \tan^{-1}y})=0\) must be true for the equation to hold.

Step 2: Analyze the First Case

The term \((1+y^2)=0\) cannot be zero because \(y^2\) is always non-negative, and thus 1 plus a non-negative number will be greater than zero. So, \(1+y^2 = 0\) does not provide a valid solution.

Step 3: Solve the Second Case

The other case to satisfy the equation is \(x-e^{2 \tan^{-1} y}=0\). This implies \(x=e^{2 \tan^{-1} y}\). The expression directly provides a functional relation between \(x\) and \(y\) which must hold.

Step 4: Check and Match Solution to Options

The relation \(x=e^{2 \tan^{-1} y}\) matches with option \((D)\) where \(x e^{2 \tan^{-1} y} = e^{\tan^{-1} y} + k\). Therefore, the solution satisfies option \((D)\) after considering arbitrary constant \(k\).

Key Concepts

Solution of Differential Equations

Differential equations are equations that involve a function and its derivatives. They are fundamental in describing many natural phenomena. To solve a differential equation means finding the unknown function that satisfies the equation.
In the exercise, we have the differential equation \((1+y^2)(x-e^{2 \tan^{-1} y}) \frac{dy}{dx} = 0\). To solve it, we need to understand the expression.
The equation can be satisfied if any of the terms leading to zero. Here, we have two parts: \((1+y^2)\) and \((x-e^{2 \tan^{-1}y})\).

• \((1+y^2)\) cannot be zero, because \(y^2\) is non-negative, making \(1+y^2 > 0\).
• Thus, we must have \(x-e^{2 \tan^{-1} y} = 0\).

This leads directly to \(x = e^{2 \tan^{-1} y}\), which relates \(x\) and \(y\) straightforwardly. Understanding these conditions helps us see that each specific case of zero leads us toward the solution.

Arctangent Function

The arctangent function, often denoted as \(\tan^{-1}\), is the inverse function of the tangent function. In our equation, it appears as part of the expression \(e^{2 \tan^{-1} y}\).
Characteristics of the arctangent function include:

• It undoes what the tangent function does. If \(\theta = \tan^{-1}(y)\), then \(\tan(\theta) = y\).
• The arctangent function has a range between \(-\pi/2\) and \(\pi/2\).

In this context, the arctangent helps create complex interactions between \(x\) and \(y\), converting multiplication and exponentiation operations. This offers a tractable way to express solutions of differential equations, enhancing our ability to find explicit expressions connecting variables.

Mathematical Problem Solving

Mathematical problem solving involves applying mathematical concepts and strategies to find the solution to a problem. Here, solving a differential equation requires logic and systematic thinking.
During problem solving:

• We first simplify equations to find possible solutions, looking at parts of the equation separately, as we did analyzing \(1+y^2\) and \(x-e^{2 \tan^{-1} y}\).
• This exercise demonstrates that not all parts might yield a solution, highlighting the importance of eliminating impossibilities, such as \(1+y^2 = 0\).
• Checking your solution against provided options, like in this case, ensures accuracy and understanding in finding \(x = e^{2 \tan^{-1} y}\).

Developing these problem-solving skills is crucial for anyone studying differential equations, as they build intuition for analyzing more complex mathematical topics.