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

Q: Wait, why didn't we use the standard integrating factor method for linear differential equations?

You absolutely could! The standard form is $\frac{dy}{dx} + \frac{1}{x}y = \log_e x$. The integrating factor would be $e^{\int (1/x) dx} = e^{\ln x} = x$. Multiplying by $x$ brings you right back to the equation we started with. Recognizing the product rule is just a 'shortcut' that saves precious time in the exam hall.

Q: Why do we choose $u = \log_e x$ and $dv = x dx$ in integration by parts?

Great question! We use the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). Since Logarithmic comes before Algebraic, we assign $u = \log_e x$. This makes $du = \frac{1}{x} dx$, which simplifies beautifully when multiplied by the integral of $dv$.

Q: Is it possible to get a non-zero constant $C$?

In this specific problem, the math works out such that $C=0$, but that is not a rule! Always include $+ C$ in your general solution. If you forget it, you might miss the entire curve's shift, leading to an incorrect particular solution.

Question

Let $y = y(x)$ be the solution of the differential equation, $x \frac{dy}{dx} + y = x \log_e x, (x > 1)$. If $2y(2) = \log_e 4 - 1$, then $y(e)$ is equal to :-

Options:

(A) $e^2/4$

(B) $e/4$

(C) $-e/2$

(D) $-e^2/2$

Sigma Percentile · Accepted Answer

### Analyzing the Hidden Symmetry

Welcome, future engineers! Today, we are going to peel back the layers of a differential equation that, at first glance, might seem like a standard, tedious problem.

Look at the equation:

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

In the world of JEE Advanced, the most powerful tool you possess is not just your ability to calculate, but your ability to **observe**. The left-hand side, $x \frac{dy}{dx} + y$, is the ghost of the product rule.

Recall that $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. If we set $u=x$ and $v=y$, then:

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

By recognizing this, we have instantly bypassed the need for an integrating factor. It is a moment of pure mathematical elegance.

### The Integration Journey

Now that we have simplified our equation to $\frac{d}{dx}(xy) = x \log_e x$, our path forward is clear: we must integrate both sides with respect to $x$. This gives us:

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

Now, we face the integral of $x \log_e x$. We turn to the trusty **ILATE rule**. Logarithmic functions take precedence over algebraic ones, so we set $u = \log_e x$ and $dv = x \, dx$.

Applying the integration by parts formula, $\int u \, dv = uv - \int v \, du$, we get:

$$xy = (\log_e x)\left(\frac{x^2}{2}\right) - \int \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) \, dx$$

Notice how the $x$ terms cancel out inside the integral. We are left with $\int \frac{x}{2} \, dx$, which integrates to $\frac{x^2}{4}$. Thus, our general solution emerges:

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

### The Detective Work

We have the general family of curves, but we need the specific one that satisfies our initial condition: $2y(2) = \log_e 4 - 1$. This is where we play detective.

By substituting $x=2$ into our general solution, we get:

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

Simplifying this, we have $2y(2) = 2 \log_e 2 - 1 + C$. Since $\log_e 4 = 2 \log_e 2$, our condition becomes:

$$2 \log_e 2 - 1 = 2 \log_e 2 - 1 + C$$

The terms cancel out perfectly, leaving us with $C=0$. We are left with our particular solution:

$$xy = \frac{x^2}{2} \log_e x - \frac{x^2}{4}$$

### The Final Victory

We are at the finish line. We need to find $y(e)$. We simply substitute $x=e$ into our particular solution:

$$e \cdot y(e) = \frac{e^2}{2} \log_e e - \frac{e^2}{4}$$

Since $\log_e e = 1$, this simplifies to:

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

Dividing both sides by $e$, we arrive at our final answer:

**$$y(e) = \frac{e}{4}$$**

You have successfully navigated the differential equation, applied the product rule, mastered integration by parts, and solved for the constant. Keep practicing, keep observing, and keep falling in love with the process!

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x lo g_{e} x, (x > 1)$ . If $2 y (2) = lo g_{e} 4 - 1$ , then $y (e)$ is equal to :-

Select Answer:

Visualized Solution

Identifying the Differential Equation

Recognizing the Exact Derivative

Rewriting and Integrating

Integration by Parts Setup

Executing the Integration

General Solution

Applying Initial Condition

Solving for Constant $C$

The Particular Solution

Calculating $y (e)$

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Hidden Symmetry

The Integration Journey

The Detective Work

The Final Victory

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