Problem 73 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(y(x)\) be the solution of the differential equation \((x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)\). Then \(y(e)\) is equal to: (a) 2 (d) 0 (b) \(2 \mathrm{e}\) (c) \(\mathrm{e}\)

Step-by-Step Solution

Answer

\(y(e) = \text{\(\mathrm{e}\)},\) option (c).

Step 1: Identify Differential Equation Type

The given differential equation is \((x \log x) \frac{dy}{dx} + y = 2x \log x\). This is a first-order linear differential equation of the form \(a(x) \frac{dy}{dx} + b(x)y = g(x)\), where \(a(x) = x \log x\), \(b(x) = 1\), and \(g(x) = 2x \log x\).

Step 2: Standard Form of Linear Equation

To solve this differential equation using an integrating factor, we first express it in the standard form: \(\frac{dy}{dx} + P(x)y = Q(x)\). Dividing the entire equation by \(x \log x\), we obtain: \(\frac{dy}{dx} + \frac{1}{x \log x}y = 2\). Here, \(P(x) = \frac{1}{x \log x}\) and \(Q(x) = 2\).

Step 3: Determine Integrating Factor

The integrating factor \(\mu(x)\) is found using \(\mu(x) = e^{\int P(x)\,dx}\). Integrating \(P(x) = \frac{1}{x \log x}\) gives \(\int \frac{1}{x \log x} \,dx = \log(\log x)\). Therefore, the integrating factor is \(\mu(x) = e^{\log(\log x)} = \log x\).

Step 4: Solve the Differential Equation

Multiply the differential equation \(\frac{dy}{dx} + \frac{1}{x \log x}y = 2\) by the integrating factor \(\log x\). This gives \((\log x)\frac{dy}{dx} + \frac{1}{x}y = 2\log x\). Recognize the left side as the derivative of \((\log x)y\), so we have \(\frac{d}{dx}((\log x)y) = 2\log x\). Integrate both sides to find \((\log x)y = x\log x + C\).

Step 5: Solve for y(x) and Particular Solution

To isolate \(y(x)\), divide by \(\log x\): \(y(x) = x + \frac{C}{\log x}\). Since the problem does not specify initial conditions, assume the simplest form where the constant \(C = 0\). Thus, \(y(x) = x\).

Step 6: Find y(e)

To evaluate \(y(e)\), use the solution found: \(y(x) = x\). Substituting \(x = e\) gives \(y(e) = e\).

Key Concepts

First-Order Linear Differential Equation

In the study of differential equations, understanding the classification of each equation is essential. A first-order linear differential equation is defined by its linearity and the presence of only the first derivative, such as \( \frac{dy}{dx} \). In general form, it can be expressed as

• \( a(x) \frac{dy}{dx} + b(x)y = g(x) \)

where \( a(x) \), \( b(x) \), and \( g(x) \) are functions of \( x \). These equations appear in many real-world scenarios, modeling systems that change at rates dependent on their current state.
In the provided exercise, the differential equation is given as \( (x \log x) \frac{dy}{dx} + y = 2x \log x \). It fits perfectly into the structure of a first-order linear differential equation, where \( a(x) = x \log x \), \( b(x) = 1 \), and \( g(x) = 2x \log x \). Once identified, the methods to solve it become clear.

Integrating Factor Method

The integrating factor method is a popular technique for solving first-order linear differential equations. This approach simplifies the problem by making the left side of the differential equation easily integrable.
The general procedure involves a few key steps:

• Convert the equation into standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \), by dividing throughout by \( a(x) \).
• Determine the integrating factor \( \mu(x) \), where \( \mu(x) = e^{\int P(x) \,dx} \).
• Multiply the entire differential equation by the integrating factor.
• Recognize and solve it as \( \frac{d}{dx} ( \mu(x)y ) = \mu(x)Q(x) \).
• Integrate both sides to find: \( \mu(x)y = \int \mu(x)Q(x) \,dx \).
• Solve for \( y(x) \).

In the example, identifying \( P(x) = \frac{1}{x \log x} \) allows us to calculate the integrating factor \( \mu(x) = \log x \). Applying this to the differential equation transforms it into a format that's straightforward to integrate.

Solution of Differential Equations

Solving differential equations effectively involves following a series of methodical steps. With first-order linear differential equations, the goal is to express \( y(x) \) in terms of \( x \). Once you've used the integrating factor, as demonstrated, you reach a point where you can isolate and determine \( y(x) \).
The solution process also includes finding particular values when specific points are substituted back into the equation. For this exercise,

• After applying the integrating factor method, we find \( y(x) = x + \frac{C}{\log x} \).
• Given the absence of initial conditions, simplify by setting the integration constant \( C = 0 \), leading to \( y(x) = x \).
• To find \( y(e) \), simply substitute \( x = e \) into the solution, resulting in \( y(e) = e \).

These steps not only provide the solution to the exercise but also form a framework for solving similar equations. This structured approach is essential for mastering differential equations.