Let y=f(x) be the solution of the differential equation dy/dx + xy/(x²-1) = (x⁶+4x)/√(1-x²), -1

Q: Wait, why did the Integrating Factor simplify to $\sqrt{1-x^2}$? I thought it would be more complicated.

That is a fantastic observation! It feels like magic, doesn't it? When we integrate $P(x) = \frac{x}{x^2-1}$, we get $\frac{1}{2} \ln(x^2-1)$. However, because our domain is $-1 < x < 1$, we must ensure the argument of the logarithm is positive. By adjusting the sign, we get $\frac{1}{2} \ln(1-x^2)$. When we raise $e$ to this power, the $\frac{1}{2}$ becomes an exponent inside the log, making it $\ln(\sqrt{1-x^2})$. Since $e^{\ln(u)} = u$, the whole thing simplifies perfectly. Trust the algebra—it is designed to be elegant!

Q: Why can we just throw away the integral of the odd function part? Is that allowed?

Absolutely! It is not just allowed; it is a powerful tool in your arsenal. The property states that for any odd function $g(x)$, the integral $\int_{-a}^{a} g(x) dx$ is always zero. Think of it geometrically: the area on the left of the y-axis is the exact negative of the area on the right. They cancel each other out perfectly. Recognizing this saves you from performing a difficult integration that would have resulted in zero anyway. Always check for symmetry in JEE problems!

Q: Is there a specific reason we chose $x = \sin\theta$ for the final integral?

Yes, there is a very specific reason! We had an integral involving $\frac{1}{\sqrt{1-x^2}}$. This is the classic 'signature' of a trigonometric substitution. Whenever you see $\sqrt{a^2-x^2}$, substituting $x = a \sin\theta$ is almost always the right path because it utilizes the identity $1 - \sin^2\theta = \cos^2\theta$. This transforms the square root into $\cos\theta$, which often cancels out the $dx$ term (which also contains $\cos\theta$). It turns a terrifying algebraic integral into a simple trigonometric one.

Question

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{6}+4x}{\sqrt{1-x^{2}}},$ $-1<x<1$ such that $f(0)=0$. If $6\int_{-1/2}^{1/2}f(x)dx=2\pi-\alpha$ then $\alpha^{2}$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of the Differential Equation

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey through the landscape of calculus. When you first look at the equation

$$\frac{dy}{dx} + \frac{xy}{x^2-1} = \frac{x^6+4x}{\sqrt{1-x^2}}$$

it is natural to feel a moment of hesitation. It looks dense, almost aggressive. But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to peel back that mask.

### Phase 1: Identifying the Skeleton

Every great solution begins with recognition. We are looking at a Linear Differential Equation (LDE). The standard form is

$$\frac{dy}{dx} + P(x)y = Q(x)$$

By simply aligning our given equation with this standard form, we identify our players:

$$P(x) = \frac{x}{x^2-1} \quad 	ext{and} \quad Q(x) = \frac{x^6+4x}{\sqrt{1-x^2}}$$

Do not rush past this step. Identifying the structure is 50% of the battle. Once you see the LDE, your brain should immediately reach for the 'Integrating Factor' (IF) toolkit. This is the key that unlocks the entire problem.

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

Now, we calculate the IF:

$$IF = e^{\int P(x) dx}$$

Substituting our $P(x)$, we face the integral $\int \frac{x}{x^2-1} dx$. Here is where the domain $-1 < x < 1$ becomes crucial. We must be careful with signs. We rewrite this as $-\int \frac{x}{1-x^2} dx$, which integrates to $\frac{1}{2} \ln(1-x^2)$.

When we place this into the exponential, we get $e^{\frac{1}{2} \ln(1-x^2)}$. Using the power rule of logarithms, this becomes $e^{\ln(\sqrt{1-x^2})}$. The exponential and the natural logarithm are inverse functions—they annihilate each other, leaving us with the beautiful, simple result:

$$IF = \sqrt{1-x^2}$$

### Phase 3: The Joy of Cancellation

Now, we write the general solution:

$$y \cdot IF = \int Q(x) \cdot IF dx$$

Watch what happens when we substitute our values:

$$y\sqrt{1-x^2} = \int \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx$$

Do you see it? The $\sqrt{1-x^2}$ in the denominator of $Q(x)$ and the $\sqrt{1-x^2}$ from our IF cancel out perfectly. The complexity vanishes. We are left with the simple integral $\int (x^6+4x) dx$. This is the moment where the problem rewards your patience. Integrating this gives us

$$y\sqrt{1-x^2} = \frac{x^7}{7} + 2x^2 + C$$

### Phase 4: The Symmetry Trap

We find $C=0$ using the initial condition $f(0)=0$. Now, we face the final hurdle:

$$I = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1-x^2}} dx$$

Many students would try to integrate this directly and get lost in the algebra. But you are a JEE student—you look for symmetry. We split the integral into two parts. The first part, involving $x^7$, is an **odd function**. Over the symmetric interval $[-\frac{1}{2}, \frac{1}{2}]$, its integral is zero. It disappears!

The second part, involving $x^2$, is an **even function**. We can simplify this to

$$2 \int_{0}^{\frac{1}{2}} \frac{2x^2}{\sqrt{1-x^2}} dx = 4 \int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$$

### Phase 5: The Trigonometric Finale

To solve $4 \int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$, we use the substitution $x = \sin	heta$. This transforms the integral into

$$4 \int_{0}^{\frac{\pi}{6}} \sin^2	heta d	heta$$

Using the identity $\sin^2	heta = \frac{1-\cos 2	heta}{2}$, we arrive at

$$2 \int_{0}^{\frac{\pi}{6}} (1-\cos 2	heta) d	heta$$

Evaluating this, we get

$$2 \left[ 	heta - \frac{\sin 2	heta}{2} ight]_{0}^{\frac{\pi}{6}} = 2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} ight) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$$

Finally, we multiply by 6 as requested:

$$6\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}ight) = 2\pi - 3\sqrt{3}$$

Comparing this to $2\pi - \alpha$, we find $\alpha = 3\sqrt{3}$. Squaring this gives us our final answer: **27**.

Enter Numerical Value:

Visualized Solution

Identify the Linear Differential Equation

Calculate the Integrating Factor ( $I F$ )

Write the General Solution

Integrate the RHS

Apply Initial Condition

Set up the Definite Integral

Apply Odd and Even Function Properties

Trigonometric Substitution

Evaluate the Trigonometric Integral

Final Calculation for $α^{2}$

The Sigma Insight: Linear Differential Equations

The Dance of the Differential Equation

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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