If the solution curve y = f(x) of the differential equation (x² - 4)y' - 2xy + 2x(4 - x²)² = 0, x

Q: Why can't we just use separation of variables?

Separation of variables requires us to isolate $y$ and $x$ on opposite sides. Here, the $y$ term is tied to $x$ in a way that makes that impossible. This is a classic linear differential equation, and the Integrating Factor method is our most powerful tool.

Q: Why do we reject $x=0$ and $x=-2\sqrt{2}$?

The problem explicitly states $x > 2$. In the JEE, domain constraints are not just suggestions; they are boundaries of reality. Any solution outside this domain is physically or mathematically irrelevant to the specific curve we are analyzing.

Q: Is there a faster way to find the maximum without expanding?

You could use the product rule on $y = (x^2-4)(12-x^2)$, but expanding it into a polynomial is often less prone to calculation errors. In the heat of the exam, choose the method that minimizes your cognitive load.

Question

If the solution curve $y = f(x)$ of the differential equation $(x^2 - 4)y' - 2xy + 2x(4 - x^2)^2 = 0, x > 2$, passes through the point $(3, 15)$, then the local maximum value of $f$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to embark on a journey through a differential equation that, at first glance, might seem like a chaotic mess of powers and variables.

But as we peel back the layers, you will see the elegance hidden within. This is the beauty of JEE Advanced mathematics: it is not about brute force; it is about finding the hidden structure.

### The Art of Transformation

We start with the given differential equation:
$$(x^2 - 4)y' - 2xy + 2x(4 - x^2)^2 = 0$$

Our first step is to bring this into the standard linear form:
$$\frac{dy}{dx} + P(x)y = Q(x)$$

To isolate $\frac{dy}{dx}$, we divide the entire equation by $(x^2 - 4)$:
$$\frac{dy}{dx} - \frac{2x}{x^2 - 4}y + \frac{2x(4 - x^2)^2}{x^2 - 4} = 0$$

Since $(4 - x^2)^2 = (x^2 - 4)^2$, we can cancel one power. The equation simplifies beautifully to:
$$\frac{dy}{dx} - \frac{2x}{x^2 - 4}y = -2x(x^2 - 4)$$

### The Integrating Factor

Now that we have the standard form, we identify $P(x) = -\frac{2x}{x^2 - 4}$. To solve this, we need the Integrating Factor ($IF$), defined as:
$$IF = e^{\int P(x) dx}$$

Let's evaluate the integral $\int -\frac{2x}{x^2 - 4} dx$. If we substitute $t = x^2 - 4$, then $dt = 2x dx$.

The integral becomes:
$$-\int \frac{1}{t} dt = -\ln|t| = -\ln(x^2 - 4)$$

Thus, the integrating factor is:
$$IF = e^{-\ln(x^2 - 4)} = e^{\ln((x^2 - 4)^{-1})} = \frac{1}{x^2 - 4}$$

### The Elegant Cancellation

The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx$. Substituting our values, we get:
$$y \cdot \left(\frac{1}{x^2 - 4}ight) = \int -2x(x^2 - 4) \cdot \left(\frac{1}{x^2 - 4}ight) dx$$

Notice the magic? The $(x^2 - 4)$ terms cancel out completely! We are left with:
$$\frac{y}{x^2 - 4} = \int -2x dx = -x^2 + C$$

This is the moment where the complexity vanishes, leaving us with a simple, elegant relationship.

### The Final Peak

We are given the point $(3, 15)$. Substituting $x=3$ and $y=15$ into the equation $\frac{y}{x^2 - 4} = -x^2 + C$:
$$\frac{15}{9 - 4} = -9 + C \Rightarrow 3 = -9 + C \Rightarrow C = 12$$

Our curve is defined by:
$$y = (x^2 - 4)(12 - x^2)$$

To find the local maximum, we expand this to $y = -x^4 + 16x^2 - 48$. Differentiating, we get:
$$y' = -4x^3 + 32x$$

Setting $y' = 0$, we find critical points at $x = 0$ and $x^2 = 8$, so $x = \pm 2\sqrt{2}$. Given the constraint $x > 2$, we accept only $x = 2\sqrt{2}$.

Checking the second derivative $y'' = -12x^2 + 32$:
At $x = 2\sqrt{2}$, $y'' = -12(8) + 32 = -64 < 0$, confirming a local maximum.

Finally, calculating the value:
$$f(2\sqrt{2}) = (8 - 4)(12 - 8) = 4 	imes 4 = 16$$

The local maximum value is **16**.

Animated Solution for Mathematics - Differential Equations: If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point $(3, 15)$ , then the local maximum value of $f$ is ......... .

Enter Numerical Value:

Visualized Solution

Analyzing the Differential Equation

Converting to Standard Linear Form

Simplifying the Equation

Identifying $P (x)$ and $I F$

Calculating the Integrating Factor

Setting up the General Solution

Integrating the RHS

Applying the Initial Condition

Finding $C$ and the Specific Curve

Finding Local Maxima: First Derivative

Identifying the Critical Point

Verification and Maximum Value

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Transformation

The Integrating Factor

The Elegant Cancellation

The Final Peak

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