Let y be the solution of the differential equation (1 - x²) dy = [xy + (x³ + 2)√(1 - x²)] dx, -1

Q: Wait, why do we divide by $(1 - x^2)dx$ at the very beginning?

Great question! We do this to transform the equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. By isolating $\frac{dy}{dx}$, we can clearly identify $P(x)$ and $Q(x)$, which is the essential first step for applying the integrating factor method.

Q: How do I know if it's a linear differential equation or something else?

Look for the structure: is there a $\frac{dy}{dx}$ term and a $y$ term, with no $y^2$ or other powers of $y$? If the equation can be rearranged into the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P$ and $Q$ are functions of $x$ only, then it is a linear differential equation. This is a very common pattern in JEE problems.

Q: Why does the integral of $2x$ vanish when we calculate the definite integral?

That is a fantastic observation! We are integrating over the symmetric interval $[-1/2, 1/2]$. The function $f(x) = 2x$ is an odd function, meaning $f(-x) = -f(x)$. The integral of any odd function over a symmetric interval $[-a, a]$ is always zero. This property is a powerful shortcut that saves time and prevents errors!

Question

Let $y$ be the solution of the differential equation $(1 - x^2) dy = [xy + (x^3 + 2)\sqrt{1 - x^2}] dx, -1 < x < 1$ and $y(0) = 0$. If $\int_{-1/2}^{1/2} \sqrt{1 - x^2} y(x) dx = k$, then $k^{-1}$ is equal to :

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex differential equation, feeling the weight of the unknown. It looks messy, but in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

We start with the given equation:
$$(1 - x^2) dy = [xy + (x^3 + 2)\sqrt{1 - x^2}] dx$$

Our first instinct is to bring order to chaos. By dividing the entire equation by $(1 - x^2) dx$, we isolate $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{xy}{1 - x^2} + \frac{(x^3 + 2)\sqrt{1 - x^2}}{1 - x^2}$$

This is the moment of clarity. We rearrange the terms to match the standard linear form:
$$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{x^3 + 2}{\sqrt{1 - x^2}}$$

### The Magic of the Integrating Factor

Now, we identify our players: $P(x) = -\frac{x}{1 - x^2}$ and $Q(x) = \frac{x^3 + 2}{\sqrt{1 - x^2}}$. The Integrating Factor ($I.F.$) is our bridge to the solution, defined as $e^{\int P(x) dx}$.

When we substitute $P(x)$, we get:
$$I.F. = e^{\int -\frac{x}{1 - x^2} dx}$$

Let's pause and breathe. This integral is simpler than it appears. By using the substitution $t = 1 - x^2$, the term $-x dx$ becomes $\frac{1}{2} dt$. The integral becomes $\frac{1}{2} \ln(1 - x^2)$, which simplifies to $\ln(\sqrt{1 - x^2})$.

Thus, our $I.F.$ is simply $\sqrt{1 - x^2}$. It is beautiful, isn't it? The complexity melts away.

### The Elegant Cancellation

Now, we apply the master formula:
$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

Substituting our values, we get:
$$y \sqrt{1 - x^2} = \int \left( \frac{x^3 + 2}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} \right) dx + C$$

Notice how the $\sqrt{1 - x^2}$ terms cancel out perfectly? This is the reward for your patience. We are left with the integral of a simple polynomial:
$$y \sqrt{1 - x^2} = \int (x^3 + 2) dx + C$$

Integrating this, we get:
$$y \sqrt{1 - x^2} = \frac{x^4}{4} + 2x + C$$

Using the initial condition $y(0) = 0$, we find that $C = 0$. Our particular solution is:
$$y \sqrt{1 - x^2} = \frac{x^4}{4} + 2x$$

### The Final Symmetry

Finally, we need to evaluate the integral $k = \int_{-1/2}^{1/2} \sqrt{1 - x^2} y(x) dx$. Substituting our expression, we have:
$$k = \int_{-1/2}^{1/2} \left( \frac{x^4}{4} + 2x \right) dx$$

Here, we use the symmetry property. The term $2x$ is an odd function, so its integral over $[-1/2, 1/2]$ is zero. The term $\frac{x^4}{4}$ is an even function, so we can write the integral as:
$$k = 2 \int_{0}^{1/2} \frac{x^4}{4} dx = \frac{1}{2} \int_{0}^{1/2} x^4 dx$$

This simplifies to:
$$k = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^{1/2} = \frac{1}{10} \left( \frac{1}{32} \right) = \frac{1}{320}$$

Since $k = \frac{1}{320}$, the final answer $k^{-1}$ is **320**. You have navigated the complexity and found the truth. Well done!

Animated Solution for Mathematics - Differential Equations: Let $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Convert to Linear Form

Identify $P (x)$ and $Q (x)$

Calculate Integrating Factor (Setup)

Calculate Integrating Factor (Execution)

Write the General Solution Formula

Substitute into General Solution

Integrate the RHS

Apply Initial Condition

Write the Particular Solution

Setup the Integral for $k$

Apply Definite Integral Properties

Evaluate the Final Integral

Calculate $k^{- 1}$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Elegant Cancellation

The Final Symmetry

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