Problem 96 - A Complete Resource Book in Mathematics for JEE Main

Question

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The solution of the equation \(x \sin \theta d \theta+\) \(\left(x^{3}-2 x^{2} \cos \theta+\cos \theta\right) d x=0\) is \(2 \cos \theta=x+c x e^{-x^{2}}\) Reason: Integrating factor \(=\frac{e^{x^{2}}}{x}\)

Step-by-Step Solution

Answer

(C) Assertion(A) is True, Reason(R) is False.

Step 1: Analyze the Given Assertion

The assertion states that the solution to the differential equation \(x \sin \theta \, d\theta +\left(x^3 - 2x^2 \cos \theta + \cos \theta\right) dx = 0\) is \(2 \cos \theta = x + cxe^{-x^2}\). We must check if this equation can be solved or verified to yield the expression mentioned using standard methods.

Step 2: Verify the Relationship in Assertion

To verify an equation as a solution, we typically isolate the variable terms and determine if the equation holds true or can be derived. However, the complexity here involving trigonometric and exponential functions suggests checking if the solution satisfies the differential equation. This usually involves integration techniques and unique manipulations including substitution.

Step 3: Examine the Given Reason

The reason provided claims that the integrating factor is \(\frac{e^{x^2}}{x}\). For a linear differential equation in the form \(Mdx + Ndy = 0\), the integrating factor can modify the differential equation to a solution. We need to calculate or identify this based on solving methods and verify if it supports the equation provided in the Assertion.

Step 4: Determine the Correctness of the Reason

We check if the integrating factor \(\frac{e^{x^2}}{x}\) is appropriate or usable in the context of the given equation for making it exact or solvable. The equation’s form does not initially suggest straightforward usage of this factor without additional context or steps demonstrating why it should fit.

Step 5: Conclude on the Validity of (A) and (R)

Based on the mathematical evaluations, \(2 \cos \theta = x + cxe^{-x^2}\) being a solution seems improbable without missteps in substantial solution transformation. The integrating factor \(\frac{e^{x^2}}{x}\) does not straightforwardly lead to clear logical resolution of the equation’s complexity. Evaluating both parts yields mismatches in applicability.

Key Concepts

Integrating Factor

Understanding the concept of an integrating factor is crucial when dealing with certain types of differential equations. An integrating factor is often used to transform a non-exact differential equation into an exact one. This means it helps in finding a solution by multiplying the entire differential equation by a certain function. This function is known as the integrating factor.

Here's the general idea:

• For a first-order linear differential equation in the form: \( \frac{dy}{dx} + P(x)y = Q(x) \),
• The integrating factor \( \mu(x) \) is given by the expression \( e^{\int P(x)\,dx} \).

The goal is to multiply the equation throughout by this integrating factor. Carried out correctly, this simplification turns it into a form that is easier to solve by integration.

Applying these principles requires careful derivation and evaluation of \( P(x) \) for an existing equation. Often, the most challenging part is identifying if an integrating factor exists and how it can be used efficiently to simplify complex relationships within the differential equation.

Linear Differential Equation

Linear differential equations are a crucial category of differential equations. In these equations, both the dependent variable and all its derivatives appear linearly. A classic example of a first-order linear differential equation is written as:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

In this equation:

• \( \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \).
• \( P(x) \) and \( Q(x) \) are functions of \( x \).

These equations can often be solved using the integrating factor method, allowing us to find a solution in terms of \( y \) and \( x \).

The simplicity of linear differential equations compared to nonlinear ones is significant. They allow for straightforward solutions under specific conditions. However, it is essential to identify whether a given equation is linear and to contextualize it properly using identified strategies like the integrating factor.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics and are often encountered in various forms within differential equations. Familiarity with these functions includes understanding sine, cosine, and tangent, among others. In many problems, these functions introduce periodic behavior into equations.

For instance, consider functions like:

• \( \sin(x) \) - gives the sine of the angle \( x \)
• \( \cos(x) \) - provides the cosine of the angle \( x \)
• \( \tan(x) \) - represents the tangent of \( x \)

Trigonometric functions play a crucial role in describing oscillations and wave-like elements in physical systems. When embedded in differential equations, they can complicate solutions but also provide insight into the periodic nature of the functions involved.

Solving differential equations with trigonometric components often requires a firm understanding of these functions' properties and how they interact with other terms in the equation. This involves using known identities and characteristics to simplify and solve equations, providing powerful tools in both theoretical and applied mathematics.