Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+[x+\log x-x \sin y] d y=0\) is (A) \(y(x+\log x)-x \cos y=c\) (B) \(y(x+\log x)+x \sin y=c\) (C) \(y(x+\log x)+x \cos y=c\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (B) \(y(x+\log x)+x \sin y=c\).

Step 1: Identify the given equation format

The given equation is \( \left[y\left(1+\frac{1}{x}\right)+\cos y\right] dx+[x+\log x-x \sin y] dy=0 \), which is a differential equation in the format \( M\,dx + N\,dy = 0 \). Here, \( M = y\left(1+\frac{1}{x}\right)+\cos y \) and \( N = x + \log x - x \sin y \).

Step 2: Check for exactness

To check if the differential equation is exact, calculate the partial derivative: \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). If they are equal, the equation is exact.For \( M \), \( \frac{\partial M}{\partial y} = \left(1+\frac{1}{x}\right) - \sin y \). For \( N \), \( \frac{\partial N}{\partial x} = 1 + \frac{1}{x} \).They are equal, meaning the equation is exact.

Step 3: Find a potential function

An exact equation has a potential function \( F(x, y) = c \). To find \( F(x, y) \), integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \), then combine the solutions considering common terms.Integrate \( M \) with respect to \( x \): \[ F(x, y) = \int \left( y\left(1+\frac{1}{x}\right) + \cos y \right) dx = yx + y\log x + x\cos y + h(y) \]Now, integrate \( N \) with respect to \( y \): \[ \int \left( x + \log x - x \sin y \right) dy = xy + y\log x - x \cos y + g(x) \]

Step 4: Match with potential solutions

Combine the information from both integrations to form the complete potential function based on common terms.From both integrations, we have: \( F(x, y) = y(x + \log x) + x \sin y \) which matches one of the given options.

Key Concepts

Exact Differential Equations

Exact differential equations are a special type of differential equation that can be written in the form \( M(x, y)\,dx + N(x, y)\,dy = 0 \). In such an equation, there exists a function \( F(x, y) \) such that \( \frac{\partial F}{\partial x} = M \) and \( \frac{\partial F}{\partial y} = N \).
This means we can find a single function, called a potential function, whose total differential is equivalent to the original differential equation. To determine whether a given differential equation is exact, we check if the partial derivative of \( M \) with respect to \( y \) is equal to the partial derivative of \( N \) with respect to \( x \).
If these derivatives are equal, the equation is exact and we can proceed to find the potential function. This approach simplifies finding solutions, making these equations more manageable.

Integration

Integration is a fundamental concept in calculus used to find functions or quantities when their rate of change is known. In the context of exact differential equations, integration is used to find the potential function, \( F(x, y) \), for which the differential equation \( M(x, y)\,dx + N(x, y)\,dy = 0 \) satisfies \( dF = 0 \).
We begin by integrating the function \( M(x, y) \) with respect to \( x \) while treating \( y \) as constant. Then, we integrate \( N(x, y) \) with respect to \( y \), treating \( x \) as constant.
These integrations give us expressions which, when combined and adjusted for consistency, yield the potential function \( F(x, y) \). This process effectively "undoes" the derivatives, allowing us to find a solution that satisfies the original exact differential equation.

Partial Derivatives

Partial derivatives involve differentiating a function with respect to one variable while keeping other variables constant. In exact differential equations, partial derivatives play a crucial role in verifying and solving the equations.
To test if a differential equation is exact, we need \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) to be equal. This condition confirms that the equation can be expressed as the total differential of some potential function \( F(x,y) \).
Calculating partial derivatives helps us break down complex expressions and find patterns or relationships between variables in multivariable functions. It is a vital tool for exploring how functions change with respect to individual variables while maintaining the impact of others.