Problem 60 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=(\tan x-y) \sec ^{2} x, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(\mathrm{y}(0)=0\) then \(y\left(-\frac{\pi}{4}\right)\) is equal to: (a) \(\mathrm{e}-2\) (b) \(\frac{1}{2}-e\) (c) \(2+\frac{1}{e}\) (d) \(\frac{1}{e}-2 \mid\)

Step-by-Step Solution

Answer

\( y\left(-\frac{\pi}{4}\right) = e - 2 \).

Step 1: Identify the Type of Differential Equation

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

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( \mu(x) = e^{\int P(x) \, dx} = e^{-\int \sec^2 x \, dx} = e^{\tan x} \).

Step 3: Solve the Linear Differential Equation

Multiply the differential equation by the integrating factor to simplify:\[ewequationn{}\begin{align*}ewequationn{}\mu(x) \frac{dy}{dx} - \mu(x)\sec^2 x \, y &= \mu(x) \tan x \sec^2 x \ewequationn{}\frac{d}{dx}[y e^{\tan x}] &= \tan x e^{\tan x} \sec^2 x.\end{align*}\]Integrate both sides with respect to \( x \):\[ y e^{\tan x} = \int \tan x e^{\tan x} \sec^2 x \, dx + C \].

Step 4: Compute the Integrals

Use substitution to solve:Let \( u = \tan x \), then \( du = \sec^2 x \, dx \). This changes the integral to:\[ \int u e^u \, du = e^u (u - 1) + C \], where we have used integration by parts.Substituting back, we have:\[ y e^{\tan x} = e^{\tan x}(\tan x - 1) + C \].

Step 5: Use the Initial Condition

Apply the given initial condition \( y(0) = 0 \) to find \( C \).Substitute \( x = 0 \) and \( y = 0 \) into the equation: \[ 0 = e^{\tan 0}(\tan 0 - 1) + C \implies C = 1 \].Therefore, the equation simplifies to:\[ y e^{\tan x} = e^{\tan x}(\tan x - 1) + 1 \].

Step 6: Solve for y using Specific x

Substitute \( x = -\frac{\pi}{4} \) into the equation:\[ y e^{\tan(-\frac{\pi}{4})} = e^{\tan(-\frac{\pi}{4})}\left(\tan(-\frac{\pi}{4}) - 1\right) + 1 \].We know \( \tan(-\frac{\pi}{4}) = -1 \), so the equation becomes:\[ y e^{-1} = e^{-1}(-1 - 1) + 1 \].

Step 7: Simplify and Solve the Equation

Simplify the equation:\[ y e^{-1} = e^{-1}(-2) + 1 \].Solving for \( y \), we have:\[ y = -2 + e \].

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is one of the simplest and most common types of differential equations you'll encounter. It is called 'first-order' because it involves the first derivative of the function. The general form of a first-order linear differential equation is:

• \( \frac{dy}{dx} + P(x) y = Q(x) \)

In this format, \( P(x) \) and \( Q(x) \) are functions of \( x \), and \( y \) is the dependent variable. The problem provided is a first-order linear differential equation with:

• \( P(x) = -\sec^2 x \)
• \( Q(x) = \tan x \sec^2 x \)

Understanding this form allows you to approach solving it with standardized methods, such as using an integrating factor.

Integrating Factor Method

The integrating factor method is a tool used to solve first-order linear differential equations. It's designed to make the equation easier to integrate. The integrating factor, \( \mu(x) \), is calculated from:

• \( \mu(x) = e^{\int P(x) \, dx} \)

For the given problem:

• \( P(x) = -\sec^2 x \)
• So, \( \mu(x) = e^{-\int \sec^2 x \, dx} = e^{-\tan x} \)

Once you find \( \mu(x) \), it is used to multiply through the entire differential equation. This converts the differential equation into an integrable form,

• \( \frac{d}{dx}[y \mu(x)] = Q(x) \mu(x) \)

Solving this form involves integrating both sides, which then allows for determining the function \( y \). Integrating factor transforms the problem into a more straightforward equation to solve.

Initial Value Problem

An initial value problem provides an initial condition to help find a particular solution to a differential equation. In our problem, the initial condition given is:

• \( y(0) = 0 \)

This condition specifies the value of \( y \) when \( x = 0 \). It is crucial because, without this condition, you could only solve the differential equation for a general solution containing an arbitrary constant \( C \).In this exercise, the initial condition is used after integrating the equation to solve for \( C \), ensuring that the solution fits this specific scenario. After applying the initial condition to the equation:

• \( y e^{\tan x} = e^{\tan x}(\tan x - 1) + C \)

you determine the specific value of \( C \) that makes \( y(0) = 0 \), leading to the complete and specific solution fitting the initial conditions provided.