If the curve y = y(x) represented by the solution of the differential equation (2xy² - y) dx + xdx =

Q: Why did we divide by $y^2$? How are we supposed to know that?

That is a fantastic question! In differential equations, we often look for an 'integrating factor' to make the equation exact. When you see terms like $x dy$ and $y dx$ together, it is a huge hint that the quotient rule or product rule is involved. Dividing by $y^2$ is a standard technique to transform the expression into the derivative of a quotient, $d(\frac{x}{y})$. It is all about recognizing the pattern!

Q: Wait, the quotient rule for $d(\frac{x}{y})$ is $\frac{y dx - x dy}{y^2}$. My equation has $x dy - y dx$. Did I mess up the sign?

You are absolutely right to be cautious! The standard quotient rule is $d(\frac{x}{y}) = \frac{y dx - x dy}{y^2}$. Since your equation has $x dy - y dx$, it is exactly the negative of that. So, $\frac{x dy - y dx}{y^2} = -d(\frac{x}{y})$. Just keep track of that negative sign, and it will cancel out perfectly when you move terms around. You are on the right track!

Q: Can I solve the differential equation by separating variables directly? Why is this 'exact differential' method better?

Separating variables is a powerful tool, but it does not always work for every equation. In this case, the 'exact differential' method is much more elegant and direct. It allows us to bypass complex substitutions and jump straight to the solution. It is like finding a shortcut through a maze—both paths might lead to the exit, but one is clearly faster and more beautiful.

Question

If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y) dx + xdx = 0$, passes through the intersection of the lines, $2x - 3y = 1$ and $3x + 2y = 8$, then $|y(1)|$ is equal to

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a differential equation; we are uncovering a hidden structure.

When you first look at the equation $(2xy^2 - y) dx + x dy = 0$, it might seem like a chaotic jumble of variables. But in the world of mathematics, chaos is often just order in disguise. Let us embark on this journey together.

### Phase 1: The Detective Work

Our first task is to bring some order to this equation. We start by expanding the terms:

$$2xy^2 dx - y dx + x dy = 0$$

Now, look closely at the terms $x dy$ and $y dx$. They are the heartbeat of this problem and are screaming for the quotient rule.

We want to group them together. Let us rearrange the equation to isolate these terms:

$$x dy - y dx = -2xy^2 dx$$

This is our first major breakthrough. We have isolated the part of the equation that looks like a differential.

### Phase 2: The Elegant Transformation

Now, we face a hurdle. The expression $x dy - y dx$ is almost the derivative of a quotient, but it is missing a denominator. We need to create a perfect differential.

By dividing the entire equation by $y^2$, we get:

$$\frac{x dy - y dx}{y^2} = -2x dx$$

Suddenly, the right side becomes a simple, integrable function of $x$. On the left side, we recognize the quotient rule:

$$d\left(\frac{x}{y}\right) = \frac{y dx - x dy}{y^2}$$

Since our expression is the negative of this, we have $-d(\frac{x}{y}) = -2x dx$, which simplifies beautifully to:

$$d\left(\frac{x}{y}\right) = 2x dx$$

This is the elegance of mathematics—a complex equation reduced to a simple, integrable form.

### Phase 3: The Geometry of the Intersection

Before we integrate, we need to know where our curve lives. The problem tells us it passes through the intersection of two lines: $2x - 3y = 1$ and $3x + 2y = 8$.

This is a classic system of linear equations. By multiplying the first by $2$ and the second by $3$, we get:

$$4x - 6y = 2$$
$$9x + 6y = 24$$

Adding these gives $13x = 26$, so $x = 2$. Substituting this back, we find $y = 1$. Our curve passes through the point $(2, 1)$.

### Phase 4: The Final Calculation

Now, we return to our differential equation. Integrating both sides of $d(\frac{x}{y}) = 2x dx$ gives us:

$$\frac{x}{y} = x^2 + C$$

Using our point $(2, 1)$, we find $C$:

$$\frac{2}{1} = 2^2 + C \Rightarrow 2 = 4 + C \Rightarrow C = -2$$

Our specific curve is $\frac{x}{y} = x^2 - 2$. Finally, to find $|y(1)|$, we set $x = 1$:

$$\frac{1}{y(1)} = 1^2 - 2 = -1$$

Thus, $y(1) = -1$. The absolute value **$|y(1)| = 1$**. We have arrived at the destination, and the journey was nothing short of spectacular.

Animated Solution for Mathematics - Differential Equations: If the curve $y = y (x)$ represented by the solution of the differential equation $(2 x y^{2} - y) d x + x d x = 0$ , passes through the intersection of the lines, $2 x - 3 y = 1$ and $3 x + 2 y = 8$ , then $∣ y (1) ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Rearranging the Equation

Creating a Perfect Differential

Identifying the Quotient Rule

Integrating Both Sides

Finding the Intersection Point

Solving for x and y

Determining the Constant C

Evaluating y(1)

Final Result

The Sigma Insight: Variable Separable Method

The Beauty of Hidden Patterns

Phase 1

Phase 2

Phase 3

Phase 4

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