Let y(x) be the solution of the differential equation (x log x) dy/dx + y = 2x log x, (x ≥ 1). Then

Q: Wait, how do I know this is a Linear Differential Equation? It looks messy.

Great question! Whenever you see a derivative $\frac{dy}{dx}$ and a $y$ term (not $y^2$ or $\sin(y)$) appearing linearly, your alarm bells should ring for a Linear Differential Equation. The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. If you can manipulate your equation into this shape, you have a guaranteed path to the solution using the Integrating Factor method.

Q: I got stuck on the integral $\int \frac{1}{x \log x} dx$. Is there a trick?

It is not a trick, but a beautiful observation! Notice that the derivative of $\log x$ is $\frac{1}{x}$. This is a classic 'function-derivative' pair. By substituting $u = \log x$, the term $\frac{1}{x} dx$ becomes $du$, transforming the integral into $\int \frac{1}{u} du$, which is simply $\log(u)$. It is elegant, isn't it?

Q: The problem didn't give me an initial condition like $y(1) = 5$. How am I supposed to find $C$?

This is a classic JEE 'trap' that tests your observation skills. Look at the domain $x \ge 1$. If you plug $x=1$ into the original equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$, the term $(x \log x)$ becomes $(1 \cdot 0) = 0$. This eliminates the derivative term entirely, leaving you with $0 + y(1) = 0$. You just found your initial condition hidden in plain sight!

Question

Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x, (x \ge 1)$. Then $y(e)$ is equal to:

Options:

(A) 2

(B) 2e

(C) e

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:
$$(x \log x) \frac{dy}{dx} + y = 2x \log x$$

At first glance, it looks intimidating. But in the world of JEE Advanced, intimidation is just a mask for structure.

### Phase 1: The Quest for Standard Form

Every Linear Differential Equation (LDE) has a 'home.' That home is the standard form:
$$\frac{dy}{dx} + P(x)y = Q(x)$$

Our equation is currently wearing a disguise. The term $(x \log x)$ is clinging to the $\frac{dy}{dx}$ like a stubborn barnacle.

To reveal the true nature of the equation, we must isolate the derivative. We divide the entire equation by $(x \log x)$.

Suddenly, the fog clears. We are left with:
$$\frac{dy}{dx} + \frac{1}{x \log x} y = 2$$

Now, we can clearly see our $P(x) = \frac{1}{x \log x}$ and our $Q(x) = 2$. The machine is starting to make sense.

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

How do we solve this? We need a bridge. In the realm of LDEs, that bridge is the Integrating Factor (I.F.).

It is defined as:
$$I.F. = e^{\int P(x) dx}$$

Let us calculate this. We need to integrate $\int \frac{1}{x \log x} dx$.

Look at the denominator. We have $\log x$, and sitting right next to it is its derivative, $\frac{1}{x}$. This is a gift from the mathematical gods.

By substituting $u = \log x$, the integral transforms into $\int \frac{1}{u} du$, which is simply $\log(u)$. Substituting back, we get $\log(\log x)$.

Now, the I.F. becomes:
$$I.F. = e^{\log(\log x)} = \log x$$

Because the exponential and logarithmic functions are inverse operations, they cancel out, leaving us with a beautifully simple $I.F. = \log x$.

### Phase 3: The General Solution

With our I.F. in hand, the general solution is just a formula away:
$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

Substituting our values, we get:
$$y \log x = \int 2 \log x dx + C$$

Integrating $\log x$ is a classic integration-by-parts exercise, yielding $x \log x - x$. Thus, our general solution is:
$$y \log x = 2(x \log x - x) + C$$

### Phase 4: The Hidden Clue

We have a constant $C$, but no initial condition. Or do we? JEE problems are rarely incomplete; they are often just subtle.

The problem specifies $x \ge 1$. If we test the boundary $x=1$ in the original equation, the term $(x \log x)$ becomes $(1 \cdot 0) = 0$.

The derivative vanishes! We are left with $y(1) = 0$. This is our hidden key.

Plugging $x=1$ and $y=0$ into our general solution:
$$0 \cdot \log 1 = 2(1 \log 1 - 1) + C$$

Since $\log 1 = 0$, we get $0 = 2(0 - 1) + C$, which simplifies to **$C = 2$**.

### The Final Victory

We have our specific solution:
$$y \log x = 2x \log x - 2x + 2$$

The question asks for $y(e)$. We substitute $x=e$. Since $\log e = 1$, the equation becomes:
$$y(e) \cdot 1 = 2e(1) - 2e + 2$$

The $2e$ terms cancel out with poetic precision, leaving us with **$y(e) = 2$**.

We have successfully decoded the machine. Remember, in mathematics, the complexity is often just a test of your patience. Keep your eyes open for the hidden clues, and the solution will always reveal itself.

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

Select Answer:

Visualized Solution

Identifying the Differential Equation

Converting to Standard Form

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

The Integrating Factor Formula

Calculating the Integrating Factor

Simplifying the Integrating Factor

General Solution Formula

Substituting into the General Solution

Integrating the Right Hand Side

The Hidden Initial Condition

Finding the Constant $C$

The Specific Solution

Evaluating at $x = e$

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be the solution of the differential equation (xlogx)dxdy​+y=2xlogx,(x≥1). Then y(e) is equal to: