Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The general solution of the differential equation, \(\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0\) (a) \(y \sqrt{\tan x}=x+c\) (b) \(y \sqrt{\cot x}=\tan x+c\) (c) \(y \sqrt{\tan x}=\cot x+c\) (d) \(y \sqrt{\cot x}=x+c\)

Step-by-Step Solution

Answer

Solution is not among the provided options.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \sin 2x \left( \frac{dy}{dx} - \sqrt{\tan x} \right) - y = 0 \). This appears to be a first-order linear differential equation because it can be rearranged to the form \( \frac{dy}{dx} - P(x) y = Q(x) \).

Step 2: Simplify the Equation

First, simplify the differential equation. By distributing \( \sin 2x \), we have: \( \sin 2x \frac{dy}{dx} - \sin 2x \sqrt{\tan x} - y = 0 \). Rearrange it to \( \sin 2x \frac{dy}{dx} = y + \sin 2x \sqrt{\tan x} \).

Step 3: Solve for \(\frac{dy}{dx}\)

The equation can be written as \( \frac{dy}{dx} = \frac{y + \sin 2x \sqrt{\tan x}}{\sin 2x} \). Write it as a linear equation, \( \frac{dy}{dx} = \frac{1}{\sin 2x}y + \sqrt{\tan x} \).

Step 4: Calculate the Integrating Factor

Recognize the linear differential equation \( \frac{dy}{dx} - \frac{1}{\sin 2x}y = \sqrt{\tan x} \). The integrating factor \( \mu(x) \) is computed as \( e^{\int P(x) dx} \), where \( P(x) = -\frac{1}{\sin 2x} \). Thus, the integrating factor is \( e^{-\int \frac{1}{\sin 2x} dx} = \sqrt{\tan x} \).

Step 5: Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor \( \sqrt{\tan x} \). This gives \( \sqrt{\tan x} \frac{dy}{dx} - \frac{1}{\sin 2x} y \sqrt{\tan x} = \tan x \).

Step 6: Solve the Differential Equation

Recognize the left-hand side after multiplication with the integrating factor as the derivative \( \frac{d}{dx} (y \sqrt{\tan x}) \). Hence, equate it to the right-hand side and integrate both sides: \( \int d(y\sqrt{\tan x}) = \int \tan x \, dx \). This results in \( y \sqrt{\tan x} = \int \tan x \, dx = \ln|\sec x| + C \).

Step 7: Rewrite the Final General Solution

The integral of \( \tan x \) gives \( \ln|\sec x| \). Thus, the general solution to the differential equation is \( y \sqrt{\tan x} = \ln|\sec x| + C \).

Key Concepts

First-Order Linear Differential Equation

Understanding a first-order linear differential equation is the first step in solving such problems. These differential equations generally take the form \(\frac{dy}{dx} + P(x)y = Q(x)\). Here:

• \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\).
• \(P(x)\) and \(Q(x)\) are functions of \(x\).
• This structure means the equation is "first-order" because it involves the first derivative of \(y\), and "linear" since \(y\) and its derivative appear to the first power without any products.

To solve such equations, we typically need to turn them into a standard format that easily allows for the application of techniques like the integrating factor. Recognizing the structure is crucial as it provides the foundation on which you can apply these solving methods.

Integrating Factor

The integrating factor is a powerful tool used to simplify the solution of linear differential equations. It helps transform the equation into an easily integrable form. Here's how it works:

• The integrating factor, \(\mu(x)\), is calculated as \(e^{\int P(x) \, dx}\), where \(P(x)\) is derived from rewriting the differential equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\).
• Multiplying the entire differential equation by this integrating factor simplifies the left-hand side into a product rule derivative. This means it can be rewritten as \(\frac{d}{dx}(\mu(x) y)\).
• This transformation is pivotal as it allows the differential to "recognize" its primitive, turning the process of solving into basic integration on both sides.

Practicing how to find and apply the integrating factor effectively will make handling first-order linear differential equations much smoother.

General Solution

The general solution of a first-order linear differential equation represents the family of functions that satisfy the equation for all \(x\). Once we've applied our integrating factor and performed the necessary integrations, this is what we achieve:

• After solving, you will typically find an expression like \(\mu(x)y = \text{Something} + C\), where \(C\) is the constant of integration.
• Here, "Something" is the integrated form of the transformed equation's right side.
• This constant \(C\) plays a crucial role, as it accounts for the various particular solutions that fit the initial conditions or specific situations of your problem.

Incorporating the constant of integration into your solution is essential because differential equations do not determine the function uniquely without it. So remember, balancing between finding expressions and interpreting them correctly is key to mastering these types of equations.