Problem 15 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x\) is the solution of the differential equation, \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{p}(x) y=\frac{2}{\pi} \operatorname{cosec} x, 0

Step-by-Step Solution

Answer

The function \( \mathrm{p}(x) \) is \( \cot x \).

Step 1: Determine the Given Differential Equation

We are given the differential equation \( \frac{\mathrm{d} y}{\mathrm{~d} x} + \mathrm{p}(x) y = \frac{2}{\pi} \operatorname{cosec} x \). The task is to find the function \( \mathrm{p}(x) \).

Step 2: Differentiate the Given y Function

Calculate \( \frac{\mathrm{d} y}{\mathrm{~d} x} \) for the function \( y = \left( \frac{2}{\pi} x - 1 \right) \operatorname{cosec} x \). Use the product rule of differentiation: if \( y = uv \), then \( \frac{dy}{dx} = u'v + uv' \). Here, let \( u = \left( \frac{2}{\pi} x - 1 \right) \) and \( v = \operatorname{cosec} x \).

Step 3: Apply the Product Rule

First, differentiate \( u \): \( u' = \frac{2}{\pi} \). Next, differentiate \( v = \operatorname{cosec} x \) which gives \( v' = -\operatorname{cosec} x \cot x \). Substitute these into the product rule to get:\[ \frac{dy}{dx} = \frac{2}{\pi} \cdot \operatorname{cosec} x + \left( \frac{2}{\pi} x - 1 \right) ( - \operatorname{cosec} x \cot x ). \]

Step 4: Simplify the Expression

Simplify to obtain:\[ \frac{dy}{dx} = \frac{2}{\pi} \operatorname{cosec} x - \left( \frac{2}{\pi} x - 1 \right) \operatorname{cosec} x \cot x. \]This simplifies further to:\[ \frac{dy}{dx} = \frac{2}{\pi} \operatorname{cosec} x - \frac{2x}{\pi} \operatorname{cosec} x \cot x + \operatorname{cosec} x \cot x. \]

Step 5: Compare with the Given Differential Equation

The equation after simplification is:\[ \frac{2}{\pi} \operatorname{cosec} x - \frac{2x}{\pi} \operatorname{cosec} x \cot x + \operatorname{cosec} x \cot x + \mathrm{p}(x) y = \frac{2}{\pi} \operatorname{cosec} x. \]Since \( y = \left( \frac{2}{\pi} x - 1 \right) \operatorname{cosec} x \), substitute y:\[ \mathrm{p}(x) \left( \frac{2}{\pi} x - 1 \right) \operatorname{cosec} x = - \frac{2x}{\pi} \operatorname{cosec} x \cot x + \operatorname{cosec} x \cot x. \]

Step 6: Determine the Function p(x)

Divide the entire equation by \( \operatorname{cosec} x \):\[ \mathrm{p}(x) \left( \frac{2}{\pi} x - 1 \right) = - \frac{2x}{\pi} \cot x + \cot x. \]Solving for \( \mathrm{p}(x) \), we have:\[ \mathrm{p}(x) = \frac{- \frac{2x}{\pi} \cot x + \cot x}{\frac{2}{\pi} x - 1}. \]Simplify and observe that \( \mathrm{p}(x) = \cot x \). Thus, \( \mathrm{p}(x) = \cot x \) is the function.

Key Concepts

Understanding the Cosecant Function

The cosecant function, \(\operatorname{cosec} x\), is a fundamental trigonometric function that is the reciprocal of the sine function. This means \(\operatorname{cosec} x = \frac{1}{\sin x}\). Understanding this relationship is crucial when solving differential equations involving trigonometric functions.
Some important properties of the cosecant function include:

• Its values are undefined wherever the sine function is zero, i.e., at integer multiples of \(\pi\).
• It is positive in the first and second quadrants of the unit circle.
• Like all trigonometric functions, it is periodic, repeating its values over a period of \(2\pi\).

Recognizing where the cosecant function is applicable helps in simplifying expressions and solving differential equations, like the one given in the exercise.

Trigonometric Functions in Differential Equations

Trigonometric functions, such as sine, cosine, and cosecant, often appear in differential equations, providing insight into oscillatory behaviors or periodic patterns.
These functions become especially important when working within specific intervals, such as \(0 < x < \frac{\pi}{2}\) for the given problem. This interval signifies a specific portion of the unit circle where direct relationships between the angles and side ratios are utilized without complications from undefined values.
In our exercise, the function \(\operatorname{cosec} x\) plays a critical role, and understanding its properties aids in the manipulation and differentiation processes that are required to solve the equation. Familiarity with these trigonometric identities allows one to simplify and reframe equations more easily, ultimately leading to the correct identification of functions like \(p(x) = \cot x\).
Always use precise definitions and properties of trigonometric functions to solve differential equations effectively.

Using the Product Rule in Differentiation

The product rule is a vital technique in calculus for differentiating functions that are products of two or more sub-functions. It states that if a function \(y = uv\), where both \(u\) and \(v\) are differentiable functions of \(x\), then the derivative is given by \(\frac{dy}{dx} = u'v + uv'\).
This rule was applied in the exercise to differentiate \(y = \left(\frac{2}{\pi} x - 1\right) \operatorname{cosec} x\). By letting \(u = \frac{2}{\pi} x - 1\) and \(v = \operatorname{cosec} x\), the product rule allows you to find the derivative \(\frac{dy}{dx}\) by:

• Computing \(u' = \frac{2}{\pi}\)
• Calculating \(v' = -\operatorname{cosec} x \cot x\)

Subsequently, substitute \(u'\) and \(v'\) into the product rule equation: \(\frac{dy}{dx} = \frac{2}{\pi} \cdot \operatorname{cosec} x + \left( \frac{2}{\pi} x - 1 \right) (- \operatorname{cosec} x \cot x)\).
Understanding and correctly applying the product rule is essential in solving complex equations involving multiple functions, as it simplifies the differentiation process.