Let y = y(x) be the solution of the differential equation x(1 - x²) dy/dx + (3x²y - y - 4x³) = 0, x

Q: Wait, why do we divide by $x(1 - x^2)$? Can't we just integrate directly?

Great question! This is a linear differential equation of the first order. The standard method to solve these is to bring them into the form $\frac{dy}{dx} + P(x)y = Q(x)$. Without this step, you cannot easily identify the Integrating Factor, which is the key to solving the equation. It is not separable, so direct integration won't work here.

Q: How do I handle the absolute value in the Integrating Factor?

The absolute value is crucial. We have $\text{I.F.} = e^{-\ln|x - x^3|} = \frac{1}{|x - x^3|}$. The problem explicitly states $x > 1$. For $x > 1$, $x^3$ is always greater than $x$, meaning $x - x^3$ is negative. Therefore, $|x - x^3| = x^3 - x$. Always check the domain provided in the question!

Q: Why does the constant $C$ appear only on one side?

When we integrate both sides of the equation $d(y \cdot \text{I.F.}) = Q(x) \cdot \text{I.F.} dx$, we technically get a constant on both sides. However, in mathematics, we combine these two constants into a single arbitrary constant $C$ on the right-hand side. It simplifies the algebra significantly without losing any generality.

Question

Let $y = y(x)$ be the solution of the differential equation $x(1 - x^2) \frac{dy}{dx} + (3x^2y - y - 4x^3) = 0, x > 1$, with $y(2) = -2$. Then $y(3)$ is equal to

Options:

(A) -18

(B) -12

(C) -6

(D) -3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the arena of differential equations. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of variables.

The equation is $x(1 - x^2) \frac{dy}{dx} + (3x^2y - y - 4x^3) = 0$, with the condition $y(2) = -2$.

It looks intimidating, doesn't it? But remember, in JEE Advanced, the most complex-looking problems often yield to the most elegant, structured approaches. Let's peel back the layers together.

### Phase 1: The Art of Rearrangement

Our first instinct is to organize. We see terms with $y$ and terms without $y$. Let's group them.

By moving the terms containing $y$ to one side and the rest to the other, we get:
$$x(1 - x^2) \frac{dy}{dx} + (3x^2 - 1)y = 4x^3$$

This is the first victory. We have transformed a chaotic expression into a recognizable structure. Now, to see the true form, we must isolate the derivative $\frac{dy}{dx}$ by dividing the entire equation by $x(1 - x^2)$.

This yields:
$$\frac{dy}{dx} + \frac{3x^2 - 1}{x(1 - x^2)} y = \frac{4x^3}{x(1 - x^2)}$$

Notice how the right-hand side simplifies beautifully:
$$\frac{dy}{dx} + \frac{3x^2 - 1}{x(1 - x^2)} y = \frac{4x^2}{1 - x^2}$$

We have arrived at the standard form of a Linear Differential Equation: $\frac{dy}{dx} + P(x)y = Q(x)$.

### Phase 2: The Integrating Factor

Now, we need the magic key: the Integrating Factor (I.F.). The formula is $	ext{I.F.} = e^{\int P(x) dx}$.

Here, $P(x) = \frac{3x^2 - 1}{x(1 - x^2)}$. Let's rewrite the denominator as $x - x^3$, so $P(x) = \frac{3x^2 - 1}{x - x^3}$.

If we look closely, the numerator is almost the derivative of the denominator. Let $u = x - x^3$, then $du = (1 - 3x^2) dx$.

Our numerator is $-(1 - 3x^2)$, so the integral becomes:
$$-\int \frac{1}{u} du = -\ln|u| = -\ln|x - x^3|$$

Thus, the I.F. is $e^{-\ln|x - x^3|} = \frac{1}{|x - x^3|}$.

Since the problem implies $x > 1$, we know $x^3 > x$, so $x - x^3$ is negative. We remove the absolute value by flipping the sign:
$$	ext{I.F.} = \frac{1}{x^3 - x} = \frac{1}{x(x^2 - 1)}$$

### Phase 3: The Integration Dance

With the I.F. in hand, the general solution is $y \cdot (	ext{I.F.}) = \int Q(x) \cdot (	ext{I.F.}) dx$.

Substituting our values, we get:
$$y \cdot \frac{1}{x(x^2 - 1)} = \int \frac{4x^2}{1 - x^2} \cdot \frac{1}{x(x^2 - 1)} dx$$

Let's simplify the integrand:
$$\frac{4x^2}{-(x^2 - 1)} \cdot \frac{1}{x(x^2 - 1)} = \frac{-4x}{(x^2 - 1)^2}$$

Now, we integrate $\int \frac{-4x}{(x^2 - 1)^2} dx$. Let $v = x^2 - 1$, so $dv = 2x dx$.

The integral becomes:
$$-2 \int v^{-2} dv = -2 \left( \frac{v^{-1}}{-1} ight) = \frac{2}{v} = \frac{2}{x^2 - 1}$$

Don't forget the constant $C$! So, the general solution is:
$$\frac{y}{x(x^2 - 1)} = \frac{2}{x^2 - 1} + C$$

### Phase 4: The Final Reveal

We are almost there. We use the initial condition $y(2) = -2$ to find $C$.

Plugging in $x=2$ and $y=-2$:
$$\frac{-2}{2(4 - 1)} = \frac{2}{4 - 1} + C \Rightarrow \frac{-2}{6} = \frac{2}{3} + C \Rightarrow -\frac{1}{3} = \frac{2}{3} + C \Rightarrow C = -1$$

Our specific solution is:
$$\frac{y}{x(x^2 - 1)} = \frac{2}{x^2 - 1} - 1$$

Finally, for $x = 3$:
$$\frac{y}{3(9 - 1)} = \frac{2}{9 - 1} - 1 \Rightarrow \frac{y}{24} = \frac{2}{8} - 1 \Rightarrow \frac{y}{24} = \frac{1}{4} - 1 = -\frac{3}{4}$$

Multiplying by 24, we get:
$$y = 24 	imes \left( -\frac{3}{4} ight) = \mathbf{-18}$$

And there it is! The journey from a messy equation to a clean, elegant result. You've mastered the process.

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

Select Answer:

Visualized Solution

Group the $y$ terms

Standard Form of L.D.E.

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

Integrating Factor Setup

Evaluate $\int P (x) d x$

Simplify the I.F.

General Solution Setup

Simplify the Integrand

Evaluate the RHS Integral

Apply Initial Condition

Find $y (3)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Group the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y terms

Standard Form of L.D.E.

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Integrating Factor Setup

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫P(x)dx