Let f: R → R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equa

Q: Why can't we just use separation of variables immediately? Isn't that the standard way to solve these?

You absolutely can! However, in competitive exams like JEE Advanced, time is your greatest asset. By performing a quick phase line analysis, you can determine the long-term behavior (the limit) without ever performing a complex partial fraction integration. It is a powerful tool to verify your analytical results and avoid calculation errors.

Q: Wait, why do we reject the equilibrium point $y=2$ and choose $y=-0.4$ as the limit?

Great question! The limit depends on the 'flow' of the system. At $y=0$, the derivative is $(2+5(0))(0-2) = -4$. Because the derivative is negative, the function $f(x)$ must decrease from its starting value of $0$. It will continue to decrease until it gets trapped by the stable equilibrium at $y=-0.4$. It cannot jump over the equilibrium points, so it is 'attracted' to $-0.4$.

Q: The official answer key says $0.4$, but my math gives $-0.4$. Did I make a mistake?

You have likely done the math perfectly! In many historical JEE papers, typos occur. Based on the equation $\frac{dy}{dx} = (2+5y)(y-2)$, the equilibrium points are indeed $y=2$ and $y=-0.4$. With $f(0)=0$, the function must decrease towards $-0.4$. If the answer key says $0.4$, it is highly probable that the original question had a sign error (perhaps $(5y-2)$ instead of $(5y+2)$). Trust your derivation!

Question

Let $f: R 	o R$ be a differentiable function with $f(0) = 0$. If $y = f (x)$ satisfies the differential equation $\frac{dy}{dx} = (2+5y)(y-2)$, then the value of $\lim_{x	o\infty} f(x)$ is ________.

Sigma Percentile · Accepted Answer

### The Dance of the Autonomous System

Welcome, future engineer. Today, we are not just solving a differential equation; we are observing the behavior of a dynamical system.

When you look at the equation

$$\frac{dy}{dx} = (2+5y)(y-2)$$

I want you to see more than just symbols. I want you to see a landscape.

This is an autonomous differential equation, which means the 'velocity' of our function $y(x)$ depends only on its position, not on the independent variable $x$. Imagine a particle moving along the $y$-axis where its speed and direction are determined entirely by its current location.

### The Equilibrium: Where Motion Ceases

Before we touch a single integral, let's find the 'steady states' of our system. These are the points where the particle stops moving entirely—the equilibrium points.

We find these by setting the derivative to zero:

$$\frac{dy}{dx} = (2+5y)(y-2) = 0$$

Solving this gives us two critical values: $y = 2$ and $y = -0.4$. These are our boundaries.

If the particle is at $y=2$ or $y=-0.4$, it stays there forever. The behavior between these points is where the story becomes interesting.

### The Flow: Determining the Path

Now, let's examine the interval between our two equilibrium points: $y \in (-0.4, 2)$. Pick any point in this range, such as $y=0$.

If we plug $y=0$ into our derivative equation, we get:

$$\frac{dy}{dx} = (2+5(0))(0-2) = -4$$

The derivative is negative, which tells us that in this entire region, the function $f(x)$ is strictly decreasing.

If our particle starts at $y=0$ (given the initial condition $f(0)=0$), it must move downwards. It will slide down the slope, approaching the stable equilibrium at $y = -0.4$. As $x$ marches toward infinity, the particle approaches this value, leading to the result:

$$\lim_{x	o\infty} f(x) = -0.4$$

### The Analytical Rigor: Confirming the Intuition

While the visual method is elegant, we must also be masters of the algebraic path. To solve this analytically, we separate the variables:

$$\int \frac{dy}{(5y+2)(y-2)} = \int dx$$

We use the method of partial fractions. Observing the identity $(5y+2) - 5(y-2) = 12$, we rewrite the integrand:

$$\frac{1}{12} \int \left( \frac{1}{y-2} - \frac{5}{5y+2} ight) dy = \int dx$$

Integrating both sides yields:

$$\frac{1}{12} \ln \left| \frac{y-2}{5y+2} ight| = x + C$$

Using our initial condition $f(0)=0$, we find that $C = \frac{1}{12} \ln | -1 | = 0$. Exponentiating both sides, we isolate the relationship:

$$\left| \frac{y-2}{5y+2} ight| = e^{12x}$$

As $x 	o \infty$, the term $e^{12x}$ grows without bound. For the left side to match this growth, the denominator $(5y+2)$ must shrink to zero.

Therefore, $5y+2 	o 0$, which confirms our limit: **$y 	o -0.4$**.

### Final Thoughts

Whether you use the geometric intuition of the phase line or the rigorous path of integration, the result remains the same. You have successfully navigated the dynamics of this system.

Remember, in JEE Advanced, the ability to visualize the behavior of a function is just as important as the ability to calculate it. Keep practicing, keep questioning, and keep falling in love with the logic behind the math.

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (y - 2)$ , then the value of $lim_{x \to \infty} f (x)$ is ________.

Enter Numerical Value:

Visualized Solution

Autonomous Differential Equation

Equilibrium Points

Phase Line Flow

Applying Initial Condition

Tracing the Solution Curve

Separating Variables

Partial Fractions

Integration

Finding the Constant

Rearranging the Equation

Evaluating the Limit

Conclusion

The Sigma Insight: Variable Separable Method

The Dance of the Autonomous System

The Equilibrium

The Flow

The Analytical Rigor

Final Thoughts

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