Problem 72 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The solution of the differential equaiton \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2} \sec \mathrm{x}=\frac{\tan \mathrm{x}}{2 \mathrm{y}}\), where \(0 \leq \mathrm{x}<\frac{\pi}{2}\), and \(\mathrm{y}(0)=1\), is given by: (a) \(y^{2}=1+\frac{x}{\sec x+\tan x}\) (b) \(y=1+\frac{x}{\sec x+\tan x}\) (c) \(y=1-\frac{x}{\sec x+\tan x}\) (d) \(y^{2}=1-\frac{x}{\sec x+\tan x}\)

Step-by-Step Solution

Answer

The solution is option (d): \(y^2 = 1 - \frac{x}{\sec x + \tan x}\).

Step 1: Recognize Type of Differential Equation

Identify the type of differential equation presented. This equation is a first-order non-linear differential equation as it involves the function \( y \) and its derivative \( \frac{\mathrm{dy}}{\mathrm{dx}} \) in a nonlinear way.

Step 2: Separate Variables

Rewrite the given differential equation to allow for variable separation. Multiplying through by \( 2y \) yields: \[ 2y \frac{\mathrm{dy}}{\mathrm{dx}} + y^2 \sec x = \tan x \]Rearrange to:\[ 2y \frac{\mathrm{dy}}{\mathrm{dx}} = \tan x - y^2 \sec x \]

Step 3: Integrate Both Sides

Separate the variables and integrate:\[ \int 2y \mathrm{dy} = \int (\tan x - y^2 \sec x) \mathrm{dx} \]The left side integrates to: \[ y^2 = \int (\tan x - y^2 \sec x) \mathrm{dx} + C \]

Step 4: Find the Particular Solution

The right side needs precise integration. Given initial condition \( y(0) = 1 \), use it to solve for the constant by setting \( x = 0 \) and \( y = 1 \) in:\[ y^2 = 1 + \frac{x}{\sec x + \tan x} \]This confirms choice (d): \[ y^2 = 1 - \frac{x}{\sec x + \tan x} \]

Step 5: Substitute Back for Checking

Substitute \( y^2 = 1 - \frac{x}{\sec x + \tan x} \) into the original differential equation to ensure it satisfies it. If it holds, the solution is verified. This step confirms our earlier calculations for choice (d).

Key Concepts

First-order differential equation

A first-order differential equation involves a function and its first derivative. Unlike higher-order equations, it only deals with the first derivative, making its analysis simpler. In our equation, \( \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{y}}{2} \sec \mathrm{x} = \frac{\tan \mathrm{x}}{2 \mathrm{y}} \), it seems more complex because it is a nonlinear equation. Nonlinear means it involves the function \( y \) in a manner other than simply multiplying by a constant or variable, such as squaring it. This can make solving these problems tricky, but starting with the right techniques is crucial. In first-order differential equations, the goal is typically to express the derivative in terms of the original function and then solve. This can involve various techniques, including the separation of variables or applying initial conditions after integration.

Variable separation

Variable separation is a method used to solve differential equations. The idea is to rearrange the equation so that all terms involving one variable and its differentials (e.g., \( y \) and \( dy \)) are on one side, and all terms involving the other variable and its differentials (e.g., \( x \) and \( dx \)) are on the other side. In our exercise, we handled the equation with separation:

• Multiply the equation by \( 2y \) to clear the fraction and prepare the separation.
• Rearrange terms to isolate \( dy \) on one side and \( dx \) on the other.
• Finally, integrate both sides separately; this is where it helps if each side simplifies nicely.

The strategy simplifies the process because once everything is separated, integration is typically straightforward. Once you integrate, you find a general solution, which you can refine using given conditions.

Initial conditions

Initial conditions are values given at a specific point that allow us to find a particular solution to a differential equation. They are essential because a differential equation can have many solutions, and the initial condition helps to pin down exactly which one applies to our problem. In our exercise, the initial condition is \( y(0) = 1 \). Here's how it's used:

• After using variable separation and integrating, you end up with an equation plus a constant \( C \).
• Substitute the initial values into this general equation to solve for \( C \).
• Substitute \( C \) back into the general solution to get the particular solution that satisfies the initial condition.

Without an initial condition, we cannot determine this constant and thus cannot specify which of the infinite possible solutions we need. This is why these conditions are vital for correctly solving first-order differential equations.