Problem 64 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution of the differential equation, \(x \frac{d y}{d x}+y=x \log _{e} \mathrm{x},(x>1) .\) If \(2 y(2)=\log _{e} 4-1\), then \(y(e)\) is equal to: [Jan. 12, 2019 (I)] (a) \(-\frac{e}{2}\) (b) \(-\frac{e^{2}}{2}\) (c) \(\frac{e}{4}\) (d) \(\frac{e^{2}}{4}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\frac{e}{4}\).

Step 1: Transform the Differential Equation

The given differential equation is \( x \frac{dy}{dx} + y = x \log_e x \). We can see that it is a first-order linear differential equation. Let's rewrite it in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) = \frac{1}{x} \) and \( Q(x) = \log_e x \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log_e x} = x \).

Step 3: Express the Solution

With the integrating factor, the solution format is \( y \times \mu(x) = \int Q(x) \times \mu(x) dx + C \). Substituting \( \mu(x) = x \) and \( Q(x) = \log_e x \), we get \( y \times x = \int x \log_e x \ dx + C \).

Step 4: Compute the Integral

The integral \( \int x \log_e x \ dx \) can be solved using integration by parts. Let \( u = \log_e x \) and \( dv = x dx \), then \( du = \frac{1}{x} dx \) and \( v = \frac{x^2}{2} \). So, \( \int x \log_e x \ dx = \frac{x^2}{2} \log_e x - \int \frac{x^2}{2} \frac{1}{x} dx = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C \).

Step 5: Write the Particular Solution

Substitute back \( y = \frac{\int x \log_e x \ dx}{x} + \frac{C}{x} \). Simplifying with the computed integral gives \( y = \frac{x}{2} \log_e x - \frac{x}{4} + \frac{C}{x} \).

Step 6: Use Initial Condition to Find C

We are given \( 2y(2) = \log_e 4 - 1 \). Substituting \( x = 2 \) into the solution form, we compute: \[ 2 \left( \frac{2}{2} \log_e 2 - \frac{2}{4} + \frac{C}{2} \right) = \log_e 4 - 1 \].\ This simplifies to \( 2(\log_e 2 - \frac{1}{2} + \frac{C}{2}) = \log_e 4 - 1 \).\ Further simplifies to \( C = 1 \).

Step 7: Find Solution at x = e

Substitute \( x = e \) into the equation derived from \( y = \frac{x}{2} \log_e x - \frac{x}{4} + \frac{C}{x} \) using \( C = 1 \):\[ y(e) = \frac{e}{2} \log_e e - \frac{e}{4} + \frac{1}{e} \].\ Simplifying gives \( \frac{e}{2} - \frac{e}{4} + \frac{1}{e} = \frac{e}{4} \).

Key Concepts

Integrating Factor

An integrating factor is a clever tool used to solve first-order linear differential equations. It helps us transform an equation into an easier one to integrate.
The integrating factor for the equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is \( \mu(x) = e^{\int P(x) \, dx} \). It effectively makes the left side of the equation into a perfect derivative, which can then be integrated directly.

• First, identify \( P(x) \), the coefficient of \( y \) in the differential equation.
• Find the integrating factor by calculating \( e^{\int P(x) \, dx} \).

For the given equation, where \( P(x) = \frac{1}{x} \), the integrating factor became \( x \).
By multiplying through with this factor, the equation is transformed to help find the solution for \( y \). It enables simplification and permits integration with ease.

Initial Condition

The initial condition is a vital part of solving differential equations. It allows us to determine the constant of integration and find a specific solution.
In the exercise, we used the initial condition \( 2y(2) = \log_e 4 - 1 \).

• After deriving the general solution of the differential equation, substitute the initial conditions.
• Plug in the given value of \( x \) (which is 2 in this case) and solve for the constant \( C \).

Once you find \( C \), substitute it back into the solution to ensure you get the correct answer for the specific conditions provided. This process helps in converting a general solution into a particular one tailored for the problem at hand. The initial condition ensures that the solution passes through a specific point in the coordinate plane.

Integration by Parts

Integration by parts is a method used to solve integrals where the integration of a product of functions is needed. It's derived from the product rule for differentiation and helps us break down complex integrals into simpler ones.
The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du.\]When dealing with the integral \( \int x \log_e x \, dx \), we chose parts as follows:

• Set \( u = \log_e x \), then \( du = \frac{1}{x} \, dx \).
• Set \( dv = x \, dx \), then \( v = \frac{x^2}{2} \).

Applying integration by parts makes tricky integrals accessible.
For our problem, it was used to evaluate \( \int x \log_e x \, dx \), resulting in \( \frac{x^2}{2} \log_e x - \frac{x^2}{4} \).
The technique effectively reduces the difficulty of the integral by facilitating its computation in terms of already known or simpler integrals, improving the complexity of solutions in integration contexts.