JEE Mathematics 2023 Question with Solution

Given:

$$x \log_e x \frac{dy}{dx} + y = x^2 \log_e x, \quad x > 1$$

and $y(2) = 2$.

Find: $y(e)$ and the correct option.

Rewrite the differential equation as

$$\frac{dy}{dx} + \frac{1}{x \log_e x} y = x$$

This is a linear differential equation of the form

$$\frac{dy}{dx} + P(x)y = Q(x)$$

where

$$P(x) = \frac{1}{x \log_e x}, \qquad Q(x) = x$$

The integrating factor is

$$\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log_e x} \, dx}$$

Let

$$u = \log_e x$$

so that

$$du = \frac{1}{x} \, dx$$

Then

$$\int \frac{1}{x \log_e x} \, dx = \int \frac{1}{u} \, du = \log_e u = \log_e(\log_e x)$$

Hence

$$\text{I.F.} = e^{\log_e(\log_e x)} = \log_e x$$

Multiplying the differential equation by $\log_e x$,

$$(\log_e x) \frac{dy}{dx} + \frac{y}{x} = x \log_e x$$

Recognize the left-hand side as

$$\frac{d}{dx}(y \log_e x) = x \log_e x$$

Integrating both sides,

$$y \log_e x = \int x \log_e x \, dx$$

Now use integration by parts for $\int x \log_e x \, dx$. Take

$$u = \log_e x, \qquad dv = x \, dx$$

Then

$$du = \frac{1}{x} \, dx, \qquad v = \frac{x^2}{2}$$

So,

$$\int x \log_e x \, dx = \frac{x^2}{2} \log_e x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C$$

Therefore,

$$y \log_e x = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C$$

Divide by $\log_e x$:

$$y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{C}{\log_e x}$$

Use the initial condition $y(2) = 2$:

$$2 = \frac{4}{2} - \frac{4}{4 \log_e 2} + \frac{C}{\log_e 2}$$

so

$$2 = 2 - \frac{1}{\log_e 2} + \frac{C}{\log_e 2}$$

Hence,

$$C = 1$$

Thus,

$$y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{1}{\log_e x}$$

Now evaluate at $x = e$. Since $\log_e e = 1$,

$$y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = \frac{e^2}{4} + 1 = \frac{4 + e^2}{4}$$

Therefore, the correct option is A.

The solution states "The Correct Option is D", but the worked calculation gives $y(e) = \frac{4 + e^2}{4}$, which matches option A.

After rewriting the equation as

$$\frac{dy}{dx} + \frac{1}{x \log_e x} y = x$$

the coefficient of $y$ is $\frac{1}{x \log_e x}$, so the integrating factor is chosen to convert the left-hand side into a product derivative.

Because

$$\frac{d}{dx}(\log_e x) = \frac{1}{x}$$

and

$$\frac{d}{dx}(y \log_e x) = (\log_e x)\frac{dy}{dx} + \frac{y}{x}$$

multiplying by $\log_e x$ creates exactly the derivative needed. That reduces the differential equation to a single integration step, followed by use of the initial condition.