Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0)\) \(=1\). Then: (a) \(y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{2}+2\) (b) \(y^{\prime}\left(\frac{\pi}{4}\right)+y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt{2}\) (c) \(y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=\sqrt{2}\) (d) \(y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}\)

Step-by-Step Solution

Answer

(a) is correct: \( y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{2}+2 \).

Step 1: Recognize the Differential Equation Form

The given differential equation is of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) = \tan x \) and \( Q(x) = 2x + x^2 \tan x \). This indicates that it's a linear first-order differential equation.

Step 2: Find the Integrating Factor

To solve the equation, we need an integrating factor, \( \mu(x) \), given by \( e^{\int P(x) \, dx} \). Here, \( \int \tan x \, dx = \log |\sec x| \), so \( \mu(x) = e^{\log |\sec x|} = \sec x \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor, \( \sec x \), to obtain: \( \sec x \frac{dy}{dx} + y(\sec x \tan x) = (2x + x^2 \tan x) \sec x \).

Step 4: Integrate Both Sides

The left side simplifies to \( \frac{d}{dx}(y \sec x) \). Thus, integrate both sides with respect to \( x \): \[ y \sec x = \int (2x \sec x + x^2 \tan x \sec x) \, dx \].

Step 5: Solve the Integral

Separate the integral into parts: \( \int 2x \sec x \, dx + \int x^2 \tan x \sec x \, dx \). These can be solved using integration by parts and substitution methods.

Step 6: Apply Initial Condition

Use the initial condition \( y(0) = 1 \) to find the constant of integration plugged back into the solution.

Step 7: Evaluate at Required Points

Once the general solution for \( y(x) \) is found, substitute \( x = \frac{\pi}{4} \) and \( x = -\frac{\pi}{4} \) to find \( y(\frac{\pi}{4}) \) and \( y(-\frac{\pi}{4}) \).

Step 8: Calculate Derivatives and Evaluate

Find the derivative \( y'(x) \) from the general expression and evaluate at \( x = \frac{\pi}{4} \) and \( x = -\frac{\pi}{4} \).

Step 9: Verify Each Option

Compute the sums and differences: (a) \( y(\frac{\pi}{4}) + y(-\frac{\pi}{4}) \), (b) \( y'(\frac{\pi}{4}) + y'(-\frac{\pi}{4}) \), (c) \( y(\frac{\pi}{4}) - y(-\frac{\pi}{4}) \), and (d) \( y'(\frac{\pi}{4}) - y'(-\frac{\pi}{4}) \). Verify which matches the given conditions.

Key Concepts

Understanding the Integrating Factor

When you encounter a linear first-order differential equation, an integrating factor can be a powerful tool for simplification and solution finding. For the equation \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is defined as \( e^{\int P(x) \, dx} \). In our example, \( P(x) \) is equal to \( \tan x \), leading to an integrating factor of \( \mu(x) = \sec x \).

This step allows you to convert the differential equation into an equation that is easier to integrate.

By multiplying the entire equation by \( \mu(x) \), you achieve a form that has an exact differential on one side, enabling straightforward integration. Think of the integrating factor as a bridge that helps capture the essence of how derivatives and integration interact in solving differential equations.

Exploring Initial Condition Problems

Initial conditions provide specific information that helps to determine the particular solution from a range of possible solutions to a differential equation. When you're given a condition like \( y(0) = 1 \), it means you're to solve the differential equation such that the function \( y \), when evaluated at \( x = 0 \), is exactly 1.

To incorporate this initial condition, you typically integrate the equation to find a general solution involving an arbitrary constant. Then, substitute the initial values to find the unique solution, giving rise to what is known as an initial condition problem.

This technique is essential because without these conditions, you would only have a family of solutions. In real-world applications, initial conditions frequently represent initial states at a given time, necessary for predicting future behavior.

What Makes a Linear Differential Equation

Linear differential equations have specific characteristics that make them unique and handy in mathematical modeling. The hallmark of a linear differential equation is that both the unknown function and its derivatives appear to the first power and are not multiplied together. For the form \( \frac{dy}{dx} + P(x)y = Q(x) \), \( P(x) \) and \( Q(x) \) are known functions of \( x \).

Linear equations are favored in many contexts due to their predictability and ease of solution. In particular, they behave nicely under operations like superposition and scaling, which means solutions can often be added together or multiplied by constants to form new solutions.

Understanding how to manipulate and solve these equations allows for broad applications, from physics and engineering to economics and beyond. Linear equations provide both the foundation and the flexibility needed for modeling change and interactions in various fields.