For the primitive integral equation y dx + y² dy = x dy; x ∈ ℝ, y 0, y(1) = 1, then y(-3) is
(a) 3

Q: Why did we divide by $y^2$? How would I know to do that in an exam?

That is a fantastic question! In differential equations, we often look for patterns that resemble the product or quotient rules. When you see $y dx$ and $x dy$ together, it is a huge hint to look for the quotient rule $\frac{y dx - x dy}{y^2}$. By dividing by $y^2$, we force the equation into a form where the numerator matches the differential of $\frac{x}{y}$. It is a technique of 'pattern matching' that comes with practice.

Q: Wait, why do we reject the negative root $y = -1$?

Great catch! Always look back at the problem statement. We were given the constraint $y > 0$. In mathematics, especially in JEE problems, constraints are not just extra information; they are essential for defining the domain of the solution. Since $y = -1$ violates the condition $y > 0$, we must discard it.

Q: Can we use the variables separable method here? It seems like it might work.

If you try to separate $x$ and $y$ directly, you will find it quite difficult because the terms are intertwined in a way that doesn't easily isolate $dx$ and $dy$. While some equations are separable, this one is designed to test your ability to recognize exact differentials. The quotient rule approach is the most efficient path to the solution.

Question

For the primitive integral equation $y dx + y^2 dy = x dy$; $x \in \mathbb{R}, y > 0, y(1) = 1$, then $y(-3)$ is

Options:

(A) 3

(B) 2

(C) 1

(D) 5

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Differentials

Welcome, fellow traveler of the mathematical landscape. Today, we are going to unravel a differential equation that, at first glance, might seem like a tangled mess of variables.

But beneath the surface lies a beautiful, elegant structure waiting to be revealed. Let us embark on this journey together.

### Phase 1: The Pattern Recognition

We start with the equation: $y dx + y^2 dy = x dy$. Our goal is to find $y(-3)$ given that $y(1) = 1$ and $y > 0$.

When you see an equation like this, do not rush to integrate. Instead, pause and observe. We have $y dx$ and $x dy$ terms, which is a classic signature of the quotient rule in reverse.

Recall that the derivative of a quotient $\frac{x}{y}$ is given by:

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

To make this pattern appear, we first rearrange our equation: $y dx - x dy + y^2 dy = 0$.

Now, the path becomes clear. If we divide the entire equation by $y^2$, we get:

$$\frac{y dx - x dy}{y^2} + \frac{y^2 dy}{y^2} = 0$$

This simplifies beautifully to:

$$\frac{y dx - x dy}{y^2} + dy = 0$$

### Phase 2: The Calculus Magic

Now, look at that first term. It is exactly the differential of $\frac{x}{y}$. Our equation has collapsed into:

$$d\left(\frac{x}{y}\right) + dy = 0$$

This is the moment of triumph! We have reduced a complex-looking equation into a simple sum of differentials.

Integrating both sides is now trivial:

$$\int d\left(\frac{x}{y}\right) + \int dy = \int 0$$

This gives us the general solution:

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

### Phase 3: Locking Down the Curve

We have a family of curves, but we need the specific one that passes through $(1, 1)$.

Substituting $x = 1$ and $y = 1$ into our equation, we get $\frac{1}{1} + 1 = C$, which means $C = 2$.

Our specific curve is defined by the equation:

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

This is the mathematical DNA of our path.

### Phase 4: The Final Destination

Our mission is to find $y$ when $x = -3$. Substituting $x = -3$ into our curve equation, we get:

$$\frac{-3}{y} + y = 2$$

To solve for $y$, we multiply by $y$ to clear the fraction:

$$-3 + y^2 = 2y \quad \Rightarrow \quad y^2 - 2y - 3 = 0$$

Factoring this quadratic, we find $(y - 3)(y + 1) = 0$. This gives us two potential roots: $y = 3$ and $y = -1$.

But remember our constraint: $y > 0$. Therefore, we must reject $y = -1$.

The only valid solution is **$y = 3$**.

Animated Solution for Mathematics - Differential Equations: For the primitive integral equation $y d x + y^{2} d y = x d y$ ; $x \in R, y > 0, y (1) = 1$ , then $y (- 3)$ is

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Grouping the Differentials

Creating an Exact Differential

Recognizing the Quotient Rule

Integrating the Equation

Applying the Initial Condition

Finding the Constant of Integration

Setting Up for the Target Value

Forming the Quadratic Equation

Solving the Quadratic Equation

Extracting the Roots

Applying the Constraint

The Sigma Insight: Linear Differential Equations

The Hidden Symmetry of Differentials

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the primitive integral equation ydx+y2dy=xdy; x∈R,y>0,y(1)=1, then y(−3) is