The general solution of the differential equation √1 + x² + y² + x²y² + xy dy/dx = 0 is (where C is

Q: Why did we factorize the square root? It looks like it would just make things more complicated.

In differential equations, our primary goal is 'Separation of Variables'. By factorizing $\sqrt{(1+x^2)(1+y^2)}$ into $\sqrt{1+x^2} \cdot \sqrt{1+y^2}$, we unlock the ability to move all $y$ terms to one side and all $x$ terms to the other. Without this step, the variables remain 'entangled', and integration becomes impossible.

Q: How do I know when to use the substitution $t = \sqrt{1+x^2}$?

Whenever you see a square root of a quadratic expression like $\sqrt{a^2+x^2}$ combined with an $x$ term in the denominator, that is a massive signal. The substitution $t = \sqrt{a^2+x^2}$ is a classic 'JEE-style' maneuver that transforms irrational algebraic functions into rational ones, which are much easier to handle.

Q: Wait, why did the negative sign disappear in the final answer?

Great catch! Remember the logarithmic property $-\ln(A/B) = \ln(B/A)$. We had a negative sign from the integration of the RHS, and by flipping the fraction inside the natural log, we absorbed that negative sign, which perfectly aligns our result with the given options.

Q: Is there a faster way to integrate $\int \frac{\sqrt{1+x^2}}{x} dx$?

You could use trigonometric substitution like $x = \tan \theta$, but that often leads to complex integrals of $\csc \theta$ or $\sec \theta$. The algebraic substitution $t = \sqrt{1+x^2}$ is generally more direct and less prone to calculation errors in the heat of an exam.

Q: Why is the constant $C$ added only at the end?

In indefinite integration, we add $C$ to represent the family of all possible solutions. While you technically get a constant from both the LHS and RHS integrals, we combine them into a single arbitrary constant $C$ because the sum or difference of two constants is still just a constant.

Question

The general solution of the differential equation $\sqrt{1 + x^2 + y^2 + x^2y^2} + xy \frac{dy}{dx} = 0$ is (where $C$ is a constant of integration)

Options:

(A) $\sqrt{1 + y^2} + \sqrt{1 + x^2} = \frac{1}{2} \log_e \left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$

(B) $\sqrt{1 + y^2} + \sqrt{1 + x^2} = \frac{1}{2} \log_e \left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right) + C$

(C) $\sqrt{1 + y^2} - \sqrt{1 + x^2} = \frac{1}{2} \log_e \left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right) + C$

(D) $\sqrt{1 + y^2} - \sqrt{1 + x^2} = \frac{1}{2} \log_e \left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of radicals and variables.

We are faced with the differential equation:

$$\sqrt{1 + x^2 + y^2 + x^2y^2} + xy \frac{dy}{dx} = 0$$

It looks intimidating, doesn't it? But remember, in the world of JEE Advanced, complexity is often just a mask for hidden elegance.

### The Art of Factorization

Our first mission is to break the 'entanglement' of $x$ and $y$. Look at the radicand: $1 + x^2 + y^2 + x^2y^2$.

If we group the first two terms and the last two terms, we get $(1 + x^2) + y^2(1 + x^2)$. Suddenly, the structure reveals itself as $(1 + x^2)(1 + y^2)$.

By factorizing the expression inside the square root, we transform our equation into:

$$\sqrt{1 + x^2} \sqrt{1 + y^2} + xy \frac{dy}{dx} = 0$$

This is the turning point! We have successfully separated the variables into a product, allowing us to isolate $y$ on one side and $x$ on the other:

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

### The Dance of Integration

Now, we integrate both sides. The left-hand side (LHS) is a classic substitution problem.

Let $u = 1 + y^2$, so $du = 2y dy$. The integral becomes:

$$\int \frac{1}{2\sqrt{u}} du = \sqrt{u} = \sqrt{1 + y^2}$$

The right-hand side (RHS) is where the real adventure begins. We have $\int \frac{\sqrt{1 + x^2}}{x} dx$.

To solve this, we use the substitution $t = \sqrt{1 + x^2}$. This implies $t^2 = 1 + x^2$, so $x^2 = t^2 - 1$ and $x dx = t dt$.

We rewrite the integral as $\int \frac{\sqrt{1 + x^2}}{x^2} \cdot x dx$, which transforms into:

$$\int \frac{t}{t^2 - 1} \cdot t dt = \int \frac{t^2}{t^2 - 1} dt$$

### The Final Simplification

To integrate $\frac{t^2}{t^2 - 1}$, we perform a simple algebraic trick: add and subtract $1$ in the numerator. This gives us:

$$\int \left( 1 + \frac{1}{t^2 - 1} \right) dt$$

The integral of $1$ is $t$, and the integral of $\frac{1}{t^2 - 1}$ is a standard form: $\frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right|$.

Substituting back $t = \sqrt{1 + x^2}$, we get:

$$\sqrt{1 + x^2} + \frac{1}{2} \ln \left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right|$$

Finally, we combine our results, remembering the negative sign from the original separation:

$$\sqrt{1 + y^2} = -\left( \sqrt{1 + x^2} + \frac{1}{2} \ln \left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right| \right) + C$$

By moving the radical to the left and using the logarithmic property $-\ln(A/B) = \ln(B/A)$, we arrive at the **elegant solution**:

$$\sqrt{1 + y^2} + \sqrt{1 + x^2} = \frac{1}{2} \ln \left| \frac{\sqrt{1 + x^2} + 1}{\sqrt{1 + x^2} - 1} \right| + C$$

Take a moment to appreciate this. We started with a chaotic differential equation and, through logical steps and algebraic intuition, arrived at a clean, structured result. Keep practicing, keep questioning, and most importantly, keep falling in love with the process!

Animated Solution for Mathematics - Differential Equations: The general solution of the differential equation $1 + x^{2} + y^{2} + x^{2} y^{2} + x y \frac{d y}{d x} = 0$ is (where $C$ is a constant of integration)

Select Answer:

Visualized Solution

Factorize the Radicand

Separate the Variables

Integrate Both Sides

Evaluate LHS Integral

Substitution for RHS

Transform RHS Integral

Simplify the Integrand

Integrate RHS Terms

Back-Substitution

Combine LHS and RHS

Apply Log Property

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Factorization

The Dance of Integration

The Final Simplification

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