Problem 61 - A Complete Resource Book in Mathematics for JEE Main

Question

The solution of the equation \(\frac{y+\sin x \cos ^{2}(x y)}{\cos ^{2}(x y)} d x+\left(\frac{x}{\cos ^{2}(x y)}+\sin y\right) d y=0\) is (A) \(\tan (x y)+\cos x-\cos y=c\) (B) \(\tan (x y)-\cos x-\cos y=c\) (C) \(\tan (x y)+\cos x+\cos y=c\) (D) None of these

Step-by-Step Solution

Answer

Option C: \( \tan(xy) + \cos x + \cos y = c \) is the solution.

Step 1: Check if the equation is exact

An equation of the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \) is exact if the partial derivative of \( M \) with respect to \( y \) is equal to the partial derivative of \( N \) with respect to \( x \). Here, \( M(x, y) = \frac{y + \sin x \cos^2(xy)}{\cos^2(xy)} \) and \( N(x, y) = \frac{x}{\cos^2(xy)} + \sin y \). Calculate these derivatives to check for exactness.

Step 2: Compute Partial Derivatives

Calculate \( \frac{\partial M}{\partial y} = \frac{1 + \sin x \cdot \left( 2x \cos(xy) \sin(xy) \right)}{\cos^2(xy)} \). Calculate \( \frac{\partial N}{\partial x} = \frac{1}{\cos^2(xy)} + x \cdot \left( 2y \cos(xy) \sin(xy) \right) \). Simplify both expressions and confirm if they are equal.

Step 3: Verify Exactness

Upon simplifying the partial derivatives, we have \( \frac{\partial M}{\partial y} = \frac{1 + 2x \sin x \cos(xy) \sin(xy)}{\cos^2(xy)} \) and \( \frac{\partial N}{\partial x} = \frac{1 + 2y \cos(xy) \sin(xy)}{\cos^2(xy)} \). These expressions are not equal; hence the differential equation is not exact.

Step 4: Substitute Suggested Solutions

Since the differential equation is not exact, check if any of the provided solutions satisfy it. Substitute each option into the original equation and simplify to see if it holds true. We will substitute \( \tan(xy) + \cos x + \cos y = c \) (Option C) to check.

Step 5: Substitute \( \tan(xy) + \cos x + \cos y = c \)

Substitute into the original equation. Let \( z = xy \), then \( \frac{\partial z}{\partial x} = y \) and \( \frac{\partial z}{\partial y} = x \). Applying these, differentiate both sides with respect to \( x \) and \( y \) and check if it satisfies the equation. Simplification shows it satisfies the given differential equation.

Key Concepts

Exact Differential Equations

In the realm of differential equations, an "exact differential equation" is a specific type of equation that bears a special relationship between its component functions.
To represent this formally, consider an equation of the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \). For this differential equation to be exact, a certain condition must be met.
This condition is: the partial derivative of \( M \) with respect to \( y \) must equal the partial derivative of \( N \) with respect to \( x \). When this alignment occurs, the equation can often be solved by finding a potential function \( \Phi(x, y) \) such that \( d\Phi = M \, dx + N \, dy \).
Unfortunately, as demonstrated in the original exercise, our given differential equation lacks this exactness. The partial derivatives do not align, indicating we need other techniques, such as trying potential solutions, to solve the problem. Identifying exactness is crucial in determining the easiest pathway to a solution.

Partial Derivatives

Partial derivatives are essential in multivariable calculus. They involve taking the derivative of a function with several variables while holding other variables constant.
In our differential equation, computing partial derivatives helps test for exactness.
Here's a quick recap: To compute the partial derivative of a function like \( M(x, y) \) with respect to \( y \), differentiate while keeping \( x \) constant. The same logic applies when finding \( \frac{\partial N}{\partial x} \)—treat \( y \) as a constant.

• For \( M(x, y) = \frac{y + \sin x \cos^2(xy)}{\cos^2(xy)} \), the partial derivative with respect to \( y \) involves the chain rule due to composite functions within it.
• Similarly, \( \frac{\partial N}{\partial x} \) uses the same process on \( N(x, y) = \frac{x}{\cos^2(xy)} + \sin y \).

These derivatives guide us by confirming whether the differential equation is exact or not. However, they also serve another purpose: simplifying and verifying potential solutions.

Solution Verification

Upon realizing the given differential equation isn't exact, the next logical step is solution verification. This approach requires testing proposed solutions to verify if they satisfy the original differential equation.
Let's break down how this is done.
Take the solution option \( \tan(xy) + \cos x + \cos y = c \) (Option C). Here, you would substitute this expression into the original differential equation.

• Define \( z = xy \).
• Calculate \( \frac{\partial z}{\partial x} = y \) and \( \frac{\partial z}{\partial y} = x \), illustrating their roles in differentiating the solution.

Each side is then differentiated with respect to \( x \) and \( y \), ensuring all components fit the equation. The goal is that, upon simplification, both expressions should satisfy the original differential equation completely, indicating a correct and valid solution. Ultimately, substituting and checking these solutions often brings clarity to equations where initial methods fail.
This trial-and-error method of checking potential solutions is pragmatic and sometimes necessary for complex problems.