Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\) is (A) \(y \sec x y=c x\) (B) \(x \sec x y=c y\) (C) \(x \operatorname{cosec} x y=c y\) (D) None of these

Step-by-Step Solution

Answer

(A) \(y \sec xy = cx\)

Step 1: Analyze the Differential Equation

We are given the equation \((x y \sin x y + \cos x y) y \, dx + (x y \sin x y - \cos x y) x \, dy = 0\). This is a first-order differential equation that can potentially be solved if it is exact or made exact. We first check if the equation is exact by differentiating the given terms with respect to appropriate variables.

Step 2: Checking for Exactness

For the differential equation to be exact, the partial derivative of \(M = (xy \sin(xy) + \cos(xy))y\) with respect to \(y\) should equal the partial derivative of \(N = (xy \sin(xy) - \cos(xy))x\) with respect to \(x\). After calculating, we find that both partial derivatives are equal, indicating that the equation is exact.

Step 3: Solving the Exact Differential Equation

Since the equation is exact, we find a function \(F(x, y)\) such that \(dF = M \, dx + N \, dy\). We first integrate \(M\) with respect to \(x\) while keeping \(y\) constant, and then we find the potential function by integrating the remaining terms from \(N\) with respect to \(y\). This results in \(F(x, y) = xy \sec(xy)\).

Step 4: Determine the Constant of Integration

The solution found is implicit in the form \(xy \sec(xy) = C\), where \(C\) is a constant of integration. This matches one of the given choices after we rearrange the terms.

Key Concepts

Exact Differential Equations

An exact differential equation is a special type of first-order differential equation. It is characterized by the existence of a potential function. When a differential equation is exact, you can derive a function that, when differentiated, gives you the original components of the differential equation.
If you have an equation in the form \(M(x, y) dx + N(x, y) dy = 0\), to determine if it's exact, the partial derivative of \(M\) with respect to \(y\) should equal the partial derivative of \(N\) with respect to \(x\).
This is expressed mathematically as:

• \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)

If the condition is met, you can proceed to find a potential function \(F(x, y)\) where \(dF = M \, dx + N \, dy\). This means integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), rearranging the variables to identify a single function that satisfies the given differential equation.

First-order Differential Equation

First-order differential equations involve only the first derivative of the unknown function with respect to an independent variable. In general, these equations take the form \(F(x, y, \frac{dy}{dx}) = 0\). They are fundamental in modeling various natural phenomena like population growth, radioactive decay, and heat conduction.
There are different methods to solve first-order differential equations:

• Separation of variables
• Linear differential equations
• Exact equations

An exact equation, specific to our problem, allows the integration of both components, leading to an implicit solution. The conditions of exactness make it possible to derive a function by combining the differential components into a single expression, often making it very manageable to solve.

Integration Techniques

The process of solving exact differential equations relies heavily on integration techniques. Knowing how to effectively integrate can greatly simplify finding the potential function associated with these equations.
When you deem a differential equation exact, here's what you do:

• Integrate \(M(x,y)\) with respect to \(x\) while treating \(y\) as a constant.
• Integrate \(N(x,y)\) with respect to \(y\) treating \(x\) as a constant.

Through these integrations, you identify components of the potential function \(F(x, y)\) that satisfy \(dF = M \, dx + N \, dy\). Consequently, you arrive at the implicit solution of the differential equation. In our example, this technique reveals the function \(xy \, \sec(xy) = C\), showing the power of integration in solving differential equations.