Let y = y(x) be the solution of the differential equation (x² + 1)y' - 2xy = (x⁴ + 2x² + 1)cos x, y(

Q: Why can't we just separate the variables and integrate directly?

This differential equation is linear, not separable. The term $-2xy$ prevents us from moving all $y$ terms to one side and $x$ terms to the other. We must use the Integrating Factor method to handle the linear structure.

Q: How do I know when to use the odd/even function property for the final integral?

Always check the limits of integration! If you see symmetric limits like $[-a, a]$, that is a huge red flag to check for symmetry. If the integrand is an odd function, the integral is zero, which saves you massive amounts of calculation time.

Q: Why did the $(x^2+1)$ term cancel out so perfectly?

That is the beauty of the Integrating Factor method! It is designed to turn the left side of the equation into the derivative of a product. When you multiply the entire equation by the I.F., the terms are mathematically engineered to simplify, leaving you with a straightforward integration.

Question

Let $y = y(x)$ be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x, y(0) = 1$. Then $\int_{-3}^3 y(x)dx$ is :

Options:

(A) 24

(B) 36

(C) 30

(D) 18

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, might look like a chaotic mess of variables and trigonometric functions. But in the world of JEE Advanced, chaos is often just order waiting to be discovered.

Let us embark on this journey to solve the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$ with the initial condition $y(0) = 1$.

### Phase 1: The Art of Pattern Recognition

The first step in any battle is reconnaissance. Look at the right-hand side: $(x^4 + 2x^2 + 1)\cos x$.

Does that polynomial look familiar? It is a perfect square in disguise! We can rewrite it as $(x^2 + 1)^2 \cos x$.

This realization is our first victory. It transforms a daunting expression into something manageable. Our goal is to force this equation into the standard linear form:

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

To do this, we divide the entire equation by $(x^2 + 1)$. The equation becomes:

$$\frac{dy}{dx} - \frac{2x}{x^2+1}y = (x^2+1)\cos x$$

Now, we have clearly identified our $P(x) = -\frac{2x}{x^2+1}$ and our $Q(x) = (x^2+1)\cos x$.

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

Now, we summon the Integrating Factor (I.F.), the secret weapon for linear differential equations. The formula is $	ext{I.F.} = e^{\int P(x) dx}$.

Substituting our $P(x)$, we get:

$$	ext{I.F.} = e^{\int -\frac{2x}{x^2+1} dx}$$

Pause here. Look at the integral. If we let $u = x^2+1$, then $du = 2x dx$. The integral becomes $-\int \frac{du}{u}$, which is $-\ln(x^2+1)$.

Thus, our I.F. is $e^{-\ln(x^2+1)}$. Using the properties of logarithms, this simplifies beautifully to:

$$	ext{I.F.} = \frac{1}{x^2+1}$$

### Phase 3: The Elegant Cancellation

Now, we multiply our standard form equation by this Integrating Factor. The left side becomes the derivative of the product $y \cdot 	ext{I.F.}$, and the right side simplifies miraculously.

We get:

$$\frac{y}{x^2+1} = \int (x^2+1)\cos x \cdot \frac{1}{x^2+1} dx + C$$

Notice how the $(x^2+1)$ terms vanish? We are left with:

$$\frac{y}{x^2+1} = \int \cos x dx + C$$

Integrating $\cos x$ gives us $\sin x$. So, $\frac{y}{x^2+1} = \sin x + C$.

Using the initial condition $y(0) = 1$, we find $1 = 0 + C$, meaning $C = 1$. Our specific solution is:

$$y(x) = (x^2+1)(\sin x + 1)$$

### Phase 4: The Symmetry Shortcut

Finally, we must evaluate $\int_{-3}^3 y(x) dx$. This is $\int_{-3}^3 (x^2+1)(\sin x + 1) dx$.

We split this into two integrals:

$$\int_{-3}^3 (x^2+1)\sin x dx + \int_{-3}^3 (x^2+1) dx$$

The first integral involves an odd function over symmetric limits, which is zero! The second integral is an even function, so it becomes:

$$2 \int_0^3 (x^2+1) dx$$

Calculating this, we get:

$$2 \left[ \frac{x^3}{3} + x ight]_0^3 = 2(9 + 3) = 24$$

And there you have it. Through pattern recognition, systematic method, and the beauty of symmetry, we have arrived at the answer: **24**.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x, y (0) = 1$ . Then $\int_{- 3}^{3} y (x) d x$ is :

Select Answer:

Visualized Solution

Identify the Differential Equation

Standard Form Conversion

Calculate the Integrating Factor

Finalize the Integrating Factor

General Solution Setup

Integrate and Find Constant $C$

Determine the Function $y (x)$

Set up the Definite Integral

Apply Odd and Even Function Properties

Evaluate the Even Function Integral

Final Calculation

The Sigma Insight: Linear Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

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