Problem 117 - A Complete Resource Book in Mathematics for JEE Main 2017

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) 0 (B) 2 (C) \(2 e\) (D) \(e\)

Step-by-Step Solution

Answer

The value of \(y(e)\) is 1.

Step 1: Identify the type of differential equation

The equation given is \[(x \log x) \frac{d y}{d x} + y = 2x \log x.\]This is a linear differential equation in standard form, where \[P(x) \cdot \frac{dy}{dx} + Q(x) \cdot y = R(x)\]with\[P(x) = x \log x, \quad Q(x) = 1, \quad R(x) = 2x \log x.\]

Step 2: Find the integrating factor

For a linear differential equation of the form \[\frac{dy}{dx} + p(x) y = q(x)\]the integrating factor is \[e^{\int p(x) dx}.\]Rewrite the equation with \[\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}.\]So, \[p(x) = \frac{1}{x \log x}.\]The integrating factor is \[e^{\int \frac{1}{x \log x} dx}.\]The integral is solved using substitution: Let \[u = \log x,\] then \[du = \frac{1}{x} dx.\]So, the integral becomes \[\int \frac{1}{u} du = \log u = \log(\log x).\]Therefore, the integrating factor is \[e^{\log(\log x)} = \log x.\]

Step 3: Multiply the differential equation by the integrating factor

Multiply each term in the equation \[\frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x}\]by the integrating factor \(\log x\):\[\log x \frac{dy}{dx} + \frac{\log x}{x \log x} y = \frac{2 \log x}{x}.\]This simplifies to:\[\log x \frac{dy}{dx} + \frac{1}{x} y = \frac{2 \log x}{x}.\]

Step 4: Recognize and solve the integrated form

The equation\[\log x \frac{dy}{dx} + \frac{1}{x}y = \frac{2 \log x}{x}\]can be rewritten as\[\frac{d}{dx}(y \cdot \log x) = \frac{2 \log x}{x}.\]Integrate both sides with respect to \(x\):\[y \cdot \log x = \int \frac{2 \log x}{x} dx.\]The right side is solved by substitution: Let \[u = \log x,\] then \[du = \frac{1}{x} dx.\]So, \[\int 2u \, du = u^2 + C = (\log x)^2 + C.\]Thus, we have:\[y \cdot \log x = (\log x)^2 + C.\]

Step 5: Solve for y

From\[y \cdot \log x = (\log x)^2 + C,\]solve for \(y\):\[y = \log x + \frac{C}{\log x}.\]

Step 6: Determine the constant using initial condition

Since there is no initial condition given for a specific value of \(y\), we will assume the simplest case where \(C = 0\) to find \(y_e\). This gives:\[y = \log x.\]

Step 7: Evaluate y(e)

Using the function \(y = \log x\), evaluate it at \(x = e\):\[y(e) = \log(e).\]Since the natural logarithm of \(e\) is 1:\[y(e) = 1.\]

Key Concepts

Linear Differential Equation

A linear differential equation is a type of differential equation in which the unknown function and its derivatives appear to the power of one (i.e., they are linear). These equations can often be written in the standard form:

• \(P(x) \cdot \frac{dy}{dx} + Q(x) \cdot y = R(x)\)

Here, \(P(x)\), \(Q(x)\), and \(R(x)\) are functions of \(x\). In our example, the equation \[(x \log x) \frac{d y}{d x} + y = 2x \log x\]is linear because both \(y\) and \(\frac{dy}{dx}\) appear linearly. Identifying the type of differential equation helps in selecting the right method for solving it.
Step 1 in our solution was about recognizing this equation as a linear differential equation, which sets the stage for choosing the appropriate techniques, such as finding an integrating factor.

Integrating Factor

An integrating factor is a function that simplifies the process of solving a linear differential equation. When you have a differential equation in the standard form like

• \[\frac{dy}{dx} + p(x) y = q(x)\]

you can use the integrating factor \(e^{\int p(x) dx}\) to make the equation easier to solve.
For our example, the equation was arranged to the form after dividing by \(x \log x\):

• \[\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}\]

Here, \( \frac{1}{x \log x} \) is our \( p(x) \). The integrating factor was found by evaluating the integral \[ \int \frac{1}{x \log x} dx \].
Transforming the equation with the integrating factor, \(\log x\), allowed for straightforward integration and helped us find the solution for \(y\). This procedure is a key step in solving linear differential equations efficiently.

Substitution Method

The substitution method is a versatile technique used to simplify complex integrals, often encountered in solving differential equations. In our example, it was used to evaluate the integral when finding the integrating factor and solving the right side of the transformed differential equation.
The substitution we made was:

• Let \(u = \log x\)
• Then, \(du = \frac{1}{x} dx\)

This substitution transformed the complex integral into a much simpler form:

• \( \int \frac{1}{u} du = \log u = \log(\log x)\)

When solving \(\int \frac{2 \log x}{x} dx\), the same substitution shows up again, making a seemingly complicated integration process much more straightforward. Understanding when and how to apply substitution is essential in solving integrals that arise in differential equations.

Natural Logarithm

The natural logarithm, denoted as \( \log x \) or sometimes \( \ln x \), is the logarithm to the base of \(e\), where \(e\) is approximately 2.71828. It is a fundamental function in calculus, particularly useful in solving aspects of differential equations.
In our given problem, the natural logarithm appears in the differential equation \[(x \log x) \frac{d y}{d x} + y = 2x \log x\].
Understanding the properties of the natural logarithm helps simplify both the equation and its integrating factor. Specifically, knowing that \( \log(e) = 1 \) simplifies calculations significantly. In the final evaluation, we use the property \( \log(e) = 1 \) to find that \(y(e) = 1\).
Mastering the use of the natural logarithm can aid in solving and understanding problems that involve growth rates, decay processes, and many natural phenomena.