Let the solution curve y = f(x) of the differential equation dy/dx + xy/(x² - 1) = (x⁴ + 2x)/√(1 - x

Q: Why does the absolute value in the Integrating Factor just vanish? Isn't $|x^2-1|$ always positive?

Great question! The absolute value is $|x^2-1|$. Since our domain is $x \in (-1, 1)$, we know $x^2 < 1$. This means $x^2 - 1$ is negative. To remove the absolute value bars, we must flip the sign to make it positive: $-(x^2 - 1) = 1 - x^2$. It is not vanishing; it is transforming based on the domain constraint. Always respect the domain!

Q: Why can we just ignore the first term in the integral? Is that mathematically legal?

It is absolutely legal! We are integrating over a symmetric interval $[-a, a]$. If a function $g(x)$ is odd, meaning $g(-x) = -g(x)$, the area on the left side of the y-axis perfectly cancels the area on the right side. Since $\frac{x^5}{5\sqrt{1-x^2}}$ is an odd function, its integral over symmetric limits is zero. It is a beautiful shortcut that saves you massive amounts of time.

Q: Why use $x = \sin \theta$ instead of integration by parts?

You *could* use integration by parts, but it would be a nightmare! The structure $\sqrt{1-x^2}$ is a massive signal to use trigonometric substitution. By setting $x = \sin \theta$, the $\sqrt{1-x^2}$ becomes $\cos \theta$, which cancels out the $dx$ term beautifully. Always look for the path of least resistance in JEE problems!

Question

Let the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}, x \in (-1, 1)$ pass through the origin. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)dx$ is equal to

Options:

(A) $\frac{\pi}{3} - \frac{1}{4}$

(B) $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$

(C) $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$

(D) $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$

Sigma Percentile · Accepted Answer

### The Detective's Guide to Differential Equations

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a mathematical investigation. When you look at a differential equation like

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

it is easy to feel overwhelmed. It looks messy, doesn't it? But remember, in the world of JEE Advanced, complexity is often a mask for elegance. Let us peel back that mask together.

### Phase 1: The Anatomy of the Equation

First, we identify our target. This is a classic First-Order Linear Differential Equation. It follows the standard form

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

By simply comparing terms, we identify our $P(x) = \frac{x}{x^2 - 1}$ and our $Q(x) = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$.

But wait! Look at the domain constraint: $x \in (-1, 1)$. This is your first clue. In JEE, constraints are never accidental.

They are the guardrails that keep you from driving off a cliff. Because $x^2 < 1$, we know that $x^2 - 1$ is negative. Keep this in your back pocket; it will save us in the next step.

### Phase 2: The Integrating Factor (The Magic Step)

To solve this, we need the Integrating Factor (I.F.), defined as $e^{\int P(x) dx}$. Substituting our $P(x)$, we get

$$e^{\int \frac{x}{x^2 - 1} dx}$$

If we multiply and divide by $2$, we get $\frac{1}{2} \int \frac{2x}{x^2 - 1} dx$. This is a standard logarithmic integral: $\frac{1}{2} \ln|x^2 - 1|$.

Now, here is where the domain constraint comes into play. We have $\ln|x^2 - 1|$. Since $x^2 < 1$, $x^2 - 1$ is negative.

To remove the absolute value, we must write it as $1 - x^2$. Thus, our I.F. becomes $e^{\ln(1 - x^2)^{1/2}}$, which simplifies beautifully to $\sqrt{1 - x^2}$. See how the complexity just melted away?

### Phase 3: The Collapse

The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. When we multiply our $Q(x)$ by our I.F., something miraculous happens.

The $\sqrt{1 - x^2}$ in the denominator of $Q(x)$ and the $\sqrt{1 - x^2}$ from our I.F. cancel out perfectly! We are left with the integral of a simple polynomial:

$$\int (x^4 + 2x) dx$$

This is the moment where you should smile. The hard part is over.

Integrating this gives us

$$y \sqrt{1 - x^2} = \frac{x^5}{5} + x^2 + C$$

Since the curve passes through the origin $(0, 0)$, we find that $C = 0$. Our function is simply

$$f(x) = \frac{x^5}{5\sqrt{1 - x^2}} + \frac{x^2}{\sqrt{1 - x^2}}$$

### Phase 4: The Symmetry Shortcut

We need to evaluate

$$\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$$

Whenever you see symmetric limits like $[-a, a]$, stop and check for odd and even functions. The first term, $\frac{x^5}{5\sqrt{1 - x^2}}$, is an odd function because replacing $x$ with $-x$ negates the entire term.

The integral of an odd function over symmetric limits is zero. Poof! It vanishes. We are left with

$$2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1 - x^2}} dx$$

This is much cleaner.

### Phase 5: The Final Push

To solve this, we use the substitution $x = \sin 	heta$. Then $dx = \cos 	heta d	heta$. When $x = 0$, $	heta = 0$. When $x = \frac{\sqrt{3}}{2}$, $	heta = \frac{\pi}{3}$.

The integral becomes

$$2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 	heta}{\cos 	heta} \cdot \cos 	heta d	heta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 	heta d	heta$$

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

$$\int_{0}^{\frac{\pi}{3}} (1 - \cos 2	heta) d	heta$$

Evaluating this, we get

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

And there you have it. A complex differential equation, tamed by domain constraints, symmetry, and trigonometric substitution. You have the tools; now go forth and conquer.

Select Answer:

Visualized Solution

Identifying the Differential Equation

Calculating the Integrating Factor

Simplifying the I.F. for the Domain

General Solution of the DE

Evaluating the Integral on the RHS

Applying the Initial Condition $(0, 0)$

Expressing the Function $f (x)$

Using Odd and Even Function Properties

Simplifying the Integral

Trigonometric Substitution

Integrating the $sin^{2} θ$ Term

Final Evaluation

The Sigma Insight: Linear Differential Equations

The Detective's Guide to Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} + \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{\frac{3}{2}}} y = 2 x exp (\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )})$ pass through the origin. Then $y (1)$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

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