Let y = y(x) be the solution of the differential equation x log_e x dy/dx + y = x² log_e x, (x 1).

Q: Why do we divide by $x \log_e x$? Can't we just integrate directly?

That is a fantastic question! If you try to integrate directly, you will find that the variables $x$ and $y$ are inextricably linked in a way that makes separation impossible. The standard form $\frac{dy}{dx} + P(x)y = Q(x)$ is a specific structure designed to allow us to use the product rule in reverse. Without this form, the 'Integrating Factor' method simply won't work. We are essentially transforming a messy equation into a perfect derivative product.

Q: I got stuck on the integral $\int x \log_e x dx$. How do I know which function is $u$ and which is $dv$?

Ah, the classic ILATE rule! Remember, I-L-A-T-E stands for Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. Since $\log_e x$ is Logarithmic and $x$ is Algebraic, Logarithmic comes first in the hierarchy. So, set $u = \log_e x$ and $dv = x dx$. This choice makes the derivative of $u$ simpler (it becomes $\frac{1}{x}$), which is exactly what we want to cancel out the $x$ in the integral!

Q: Why do we need the condition $y(2) = 2$?

Differential equations give us a 'family' of solutions, represented by the constant $C$. This is the 'general solution.' But in physics and engineering, we usually need a specific path. The condition $y(2) = 2$ acts like a pin on a map, allowing us to lock in the exact value of $C$ and find the 'particular solution' that fits our specific scenario. Without it, we would have an infinite number of curves; with it, we have the one true path.

Question

Let $y = y(x)$ be the solution of the differential equation $x \log_e x \frac{dy}{dx} + y = x^2 \log_e x, (x > 1)$. If $y(2) = 2$, then $y(e)$ is equal to

Options:

(A) $\frac{4 + e^2}{4}$

(B) $\frac{1 + e^2}{4}$

(C) $\frac{2 + e^2}{2}$

(D) $\frac{1 + e^2}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of linear differential equations.

When you first look at the equation $x \log_e x \frac{dy}{dx} + y = x^2 \log_e x$, it might seem like a chaotic jumble of variables. In the world of JEE Advanced, chaos is just order waiting to be discovered.

### Phase 1: The Art of Standardization

Every great journey begins with preparation. Our differential equation is currently cluttered with a coefficient $x \log_e x$ sitting in front of $\frac{dy}{dx}$.

To solve this, we must bring the equation into the **Standard Form** of a linear differential equation: 
$$\frac{dy}{dx} + P(x)y = Q(x)$$

By dividing the entire equation by $x \log_e x$, we perform a surgical strike on the complexity. The equation transforms into:
$$\frac{dy}{dx} + \frac{1}{x \log_e x} y = x$$

Suddenly, the fog lifts. We can clearly identify our components: $P(x) = \frac{1}{x \log_e x}$ and $Q(x) = x$.

### Phase 2: The Magic of the Integrating Factor

Now, we introduce the most powerful tool in our arsenal: the Integrating Factor (**I.F.**). The formula is defined as:
$$I.F. = e^{\int P(x) dx}$$

When we plug in our $P(x)$, we get $e^{\int \frac{1}{x \log_e x} dx}$. If we use the substitution $u = \log_e x$, then $du = \frac{1}{x} dx$.

The integral becomes $\int \frac{1}{u} du$, which is simply $\log_e u$. Substituting back, we get $\log_e(\log_e x)$.

Now, the beauty of the exponential function kicks in. We have $e^{\log_e(\log_e x)}$. Because $e$ and $\log_e$ are inverse functions, they cancel each other out, leaving us with a beautifully simple **I.F.**:
$$I.F. = \log_e x$$

### Phase 3: The Dance of Integration by Parts

With our **I.F.** in hand, the general solution is within our grasp:
$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

Substituting our values, we get $y \log_e x = \int x \log_e x dx + C$. Now, we must tackle the integral on the right using the **ILATE** rule, setting $u = \log_e x$ and $dv = x dx$.

Applying the integration by parts formula $\int u dv = uv - \int v du$, we get:
$$\int x \log_e x dx = \frac{x^2}{2} \log_e x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$$

Simplifying this, we get $\frac{x^2}{2} \log_e x - \frac{1}{2} \int x dx$, which results in:
$$\int x \log_e x dx = \frac{x^2}{2} \log_e x - \frac{x^2}{4}$$

### Phase 4: The Final Reveal

Our general solution is:
$$y \log_e x = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C$$

We are given the condition $y(2) = 2$. By substituting $x=2$ and $y=2$, we find:
$$2 \log_e 2 = \frac{4}{2} \log_e 2 - \frac{4}{4} + C$$

The $2 \log_e 2$ terms cancel out, leaving us with $0 = -1 + C$, which means **$C = 1$**.

Finally, we find $y(e)$. Plugging $x=e$ into our particular solution and knowing that $\log_e e = 1$:
$$y(1) = \frac{e^2}{2}(1) - \frac{e^2}{4} + 1$$

This simplifies to our final answer:
**$$y = \frac{e^2 + 4}{4}$$**

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Select Answer:

Visualized Solution

Standardizing the Equation

Identifying $P (x)$ and $Q (x)$

Calculating the Integrating Factor (I.F.)

Solving the Integral for I.F.

Finalizing the I.F.

Writing the General Solution

Integration by Parts

Evaluating the Integral

The General Solution

Applying Condition $y (2) = 2$

The Particular Solution

Finding $y (e)$

Final Calculation

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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