Solution of the differential equation ydx + (x + x²y)dy = 0 is (a) log y = Cx (b) -1/xy + log y = C

Q: Wait, why did we divide by $(xy)^2$? How would I know to do that in an exam?

That is a brilliant question. In differential equations, we often look for a 'separating factor.' After substituting $d(xy)$, we had $d(xy) + x^2ydy = 0$. We want the second term to be a function of $(xy)$ so it matches the first term. By dividing by $(xy)^2$, the $x^2$ terms cancel out, leaving us with a function of $y$ only, which is perfectly integrable. It is a classic pattern-matching technique—always look for what needs to be 'canceled' to make the variables independent.

Q: Can I solve this by rearranging it into $\frac{dy}{dx} = -\frac{y}{x+x^2y}$? Is that easier?

You absolutely can! That is a valid approach. However, if you try to solve it as a standard first-order differential equation, you will find it is a Bernoulli equation or requires a specific substitution. It is much more complex than the 'exact differential' method we used. In JEE, recognizing the exact differential $d(xy)$ is the 'pro' move that saves you five minutes of tedious algebra.

Q: Why is the integral of $\frac{1}{y} dy$ just $\log y$ and not something else?

Great catch! Technically, it is $\log|y|$. In most JEE problems, we assume the variables are positive within the domain of the function, so we write $\log y$. It is the fundamental integral of the reciprocal function. Never forget that constant of integration $C$—it is the most common place students lose marks!

Question

Solution of the differential equation $ydx + (x + x^2y)dy = 0$ is

Options:

(A) $\log y = Cx$

(B) $-\frac{1}{xy} + \log y = C$

(C) $\frac{1}{xy} + \log y = C$

(D) $-\frac{1}{xy} = C$

Sigma Percentile · Accepted Answer

### The Art of Observation: Unlocking the Differential Equation

Welcome, future engineers! Today, we are going to dismantle a differential equation that looks like a chaotic mess but is actually a masterpiece of mathematical elegance. The equation is:

$$ydx + (x + x^2y)dy = 0$$

When you first see this on a JEE Advanced paper, your instinct might be to panic. The variables $x$ and $y$ are tangled together, and it does not look like a standard linear equation.

But here is the secret: **Mathematics is not about calculation; it is about observation.**

### Phase 1: The Hidden Pattern

Let's expand the brackets. By distributing the $dy$, we get:

$$ydx + xdy + x^2ydy = 0$$

Now, stop and look at the first two terms: $ydx + xdy$. Does that ring a bell? It should! This is the classic product rule for differentiation.

We know that the differential of a product $xy$ is given by:

$$d(xy) = xdy + ydx$$

This is a beautiful, exact differential. By recognizing this, we have instantly simplified our equation to:

$$d(xy) + x^2ydy = 0$$

### Phase 2: The Strategic Pivot

We have made progress, but we are not home yet. We have $d(xy)$ in the first term, but the second term, $x^2ydy$, is still problematic. We need to separate the variables so we can integrate.

The goal is to make the second term a function of $y$ or a function of $(xy)$. Here is the 'JEE trick': divide the entire equation by $(xy)^2$.

Why? Because we want to force the second term to simplify. Let's see what happens:

$$\frac{d(xy)}{(xy)^2} + \frac{x^2ydy}{(xy)^2} = 0$$

Look at that second term: $\frac{x^2ydy}{x^2y^2}$. The $x^2$ terms cancel out perfectly, and one $y$ cancels out, leaving us with $\frac{dy}{y}$.

Our equation is now:

$$\frac{d(xy)}{(xy)^2} + \frac{dy}{y} = 0$$

### Phase 3: The Final Integration

Now, the variables are separated. We can integrate both sides with confidence:

$$\int \frac{d(xy)}{(xy)^2} + \int \frac{dy}{y} = C$$

For the first integral, treat $(xy)$ as a single variable, say $u$. We are simply integrating $u^{-2} du$, which gives us $-u^{-1}$, or $-\frac{1}{xy}$.

The second integral is the standard natural logarithm, $\log y$. Putting it all together, we get:

$$-\frac{1}{xy} + \log y = C$$

### The Takeaway

This problem is a perfect example of why we study differential equations. It is not just about following a recipe; it is about training your eyes to spot the product rule, the chain rule, and the exact differentials hidden in the noise.

When you face a problem like this, take a breath, expand the terms, and look for the pattern. You have the tools; you just need to trust your intuition.

Keep practicing, stay curious, and remember—**every complex equation is just a simple one waiting to be revealed.**

Animated Solution for Mathematics - Differential Equations: Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

Select Answer:

Visualized Solution

Analyze the Differential Equation

Expand the Terms

Identify Exact Differential

Substitute $d (x y)$

Strategy for Variable Separation

Divide by $(x y)^{2}$

Simplify the Terms

Integrate Both Sides

Evaluate First Integral

Evaluate Second Integral

Final Solution

The Sigma Insight: Variable Separable Method

The Art of Observation

Phase 1

Phase 2

Phase 3

The Takeaway

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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; } Solution of the differential equation ydx+(x+x2y)dy=0 is