Let y = y(x) be a differentiable function in the interval (0, ∞) such that y(1) = 2, and lim t→x (t²

Q: Wait, why do we differentiate with respect to $t$ and not $x$? This is so confusing!

I completely understand why that feels counter-intuitive! In calculus, we are trained to look for $x$ as the primary variable. However, look at the limit notation: $\lim_{t \to x}$. The variable that is 'moving' or 'changing' is $t$. As $t$ approaches $x$, $x$ is effectively a fixed point or a constant parameter. Therefore, when we apply L'Hopital's Rule, we must differentiate with respect to the variable of the limit, which is $t$. Treat $x$ as a constant, and the fog will clear immediately.

Q: How on earth was I supposed to know to divide by $x^4$? Is there a trick?

That is a fantastic question! It is not a random guess; it is pattern recognition. When you see an expression like $x^2 y' - 2x y$, look at the coefficients. The derivative of $x^2$ is $2x$. This structure screams 'Quotient Rule'. The Quotient Rule for $\frac{u}{v}$ is $\frac{v u' - u v'}{v^2}$. Here, if $v = x^2$, then $v^2 = x^4$. By dividing by $x^4$, we force the expression to match the structure of the derivative of a quotient. It is a beautiful piece of mathematical engineering!

Q: Can I use the standard integrating factor method $e^{\int P(x) dx}$ instead of the quotient rule trick?

Absolutely! You are mathematically correct to do so. The integrating factor method is the 'universal' tool for linear differential equations of the form $y' + P(x)y = Q(x)$. However, in JEE Advanced, time is your most precious resource. Recognizing the quotient rule pattern allows you to bypass the integration of the integrating factor, saving you precious minutes. Both methods lead to the same destination, but one is a scenic route and the other is a high-speed express train.

Question

Let $y = y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1) = 2$, and $\lim_{t 	o x} \frac{t^2 y(x) - x^2 y(t)}{x - t} = 3$ for each $x > 0$. Then $2y(2)$ is equal to

Options:

(A) 23

(B) 27

(C) 12

(D) 18

Sigma Percentile · Accepted Answer

### The Hidden Geometry of the Limit

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are decoding a secret message hidden in the language of calculus.

When you first look at the expression $\lim_{t 	o x} \frac{t^2 y(x) - x^2 y(t)}{x - t} = 3$, it might look like a standard limit problem. But look closer; it is a gateway to a differential equation. The beauty of this problem lies in how it forces us to bridge the gap between limits and differential equations.

### Phase 1: The Indeterminate Trap

Whenever you see a limit involving a function $y(x)$ and a variable $t$, your first instinct should always be to test the waters. What happens if we substitute $t = x$?

The numerator becomes $x^2 y(x) - x^2 y(x)$, which is $0$. The denominator becomes $x - x$, which is $0$. We have hit a $\frac{0}{0}$ indeterminate form.

In the JEE exam hall, this is your signal. It is not a dead end; it is an invitation to use L'Hopital's Rule. We must differentiate with respect to $t$, treating $x$ as a constant parameter.

Differentiating the numerator with respect to $t$:

$$\frac{d}{dt} [t^2 y(x) - x^2 y(t)] = 2t y(x) - x^2 y'(t)$$

Differentiating the denominator with respect to $t$:

$$\frac{d}{dt} [x - t] = -1$$

The limit now becomes:

$$\lim_{t 	o x} \frac{2t y(x) - x^2 y'(t)}{-1} = 3$$

### Phase 2: The Birth of the Differential Equation

Now that the indeterminate form is resolved, we can safely let $t$ approach $x$. The expression simplifies beautifully:

$$\frac{2x y(x) - x^2 y'(x)}{-1} = 3$$

Multiplying by $-1$ and rearranging, we arrive at a first-order linear differential equation:

$$x^2 y'(x) - 2x y(x) = 3$$

This is the heart of the problem. We have successfully moved from a limit to a differential equation.

### Phase 3: The Elegance of the Quotient Rule

Here is where we move from 'solving' to 'mastering'. You could use the standard integrating factor method, but look at the left side: $x^2 y'(x) - 2x y(x)$. It is the numerator of the quotient rule for $\frac{y(x)}{x^2}$.

Recall that:

$$\frac{d}{dx} \left( \frac{y}{x^2} ight) = \frac{x^2 y' - y(2x)}{(x^2)^2} = \frac{x^2 y' - 2x y}{x^4}$$

To make our equation match this, we divide the entire equation by $x^4$:

$$\frac{x^2 y'(x) - 2x y(x)}{x^4} = \frac{3}{x^4}$$

This transforms the left side into an exact derivative:

$$\frac{d}{dx} \left( \frac{y(x)}{x^2} ight) = 3x^{-4}$$

### Phase 4: The Final Integration

We are now in the home stretch. We integrate both sides with respect to $x$:

$$\int \frac{d}{dx} \left( \frac{y(x)}{x^2} ight) dx = \int 3x^{-4} dx$$

$$\frac{y(x)}{x^2} = 3 \left( \frac{x^{-3}}{-3} ight) + C$$

$$\frac{y(x)}{x^2} = -\frac{1}{x^3} + C$$

We are given the initial condition $y(1) = 2$. Let us use this to find our constant $C$:

$$\frac{2}{1^2} = -\frac{1}{1^3} + C \Rightarrow 2 = -1 + C \Rightarrow C = 3$$

With $C$ found, our function is revealed:

$$\frac{y(x)}{x^2} = -\frac{1}{x^3} + 3$$

Multiplying by $x^2$, we get:

$$y(x) = -\frac{1}{x} + 3x^2$$

### The Conclusion

The problem asks for $2y(2)$. Let us calculate $y(2)$:

$$y(2) = -\frac{1}{2} + 3(2^2) = -0.5 + 12 = 11.5$$

Finally, the result is:

**$2y(2) = 2 	imes 11.5 = 23$**

Look at what you have achieved. You navigated a limit, identified a differential equation, recognized a quotient rule pattern, and solved for the function. This is the essence of JEE Advanced mathematics—not just calculation, but the ability to see the structure beneath the symbols.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $lim_{t \to x} \frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t} = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

Select Answer:

Visualized Solution

Analyzing the Limit Expression

Applying L'Hopital's Rule

Forming the Differential Equation

Recognizing the Quotient Rule Pattern

Integrating to Find the General Solution

Using Initial Condition to Find $C$

Finding the Specific Function $y (x)$

Calculating the Final Answer

The Sigma Insight: Linear Differential Equations

The Hidden Geometry of the Limit

Phase 1

Phase 2

Phase 3

Phase 4

The Conclusion

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