The solution of the differential equation (x² + y²) dx - 5xy dy = 0, y(1) = 0, is : (1) |x² - 2y²|⁶

Q: Why do we use $y = vx$? Can't we just solve it directly?

That is a fantastic question! If you try to solve the original equation directly, you will find that the variables $x$ and $y$ are 'entangled'—they are inseparable. The substitution $y = vx$ acts like a mathematical prism; it separates the variables into $v$ and $x$ so we can integrate them independently. It is the standard key to unlocking the door of homogeneous equations.

Q: I got stuck on the integration. Why is there a $-5/8$ coefficient?

That comes from the method of substitution in integration. Since the derivative of the denominator $(1 - 4v^2)$ is $-8v$, we need to create that $-8v$ in the numerator to use the $\int \frac{1}{u} du$ rule. We multiply and divide by $-8/5$ to balance the equation, which leaves us with that specific constant. It is just a matter of balancing the scales!

Q: Wait, why do we reject the negative root or handle the modulus like that?

In differential equations, the modulus $| \dots |$ appears naturally because of the $\ln$ function. When we exponentiate both sides to remove the logarithm, the constant of integration $C$ absorbs the sign. We don't 'reject' anything; we simply use the initial condition $y(1) = 0$ to determine the specific value of the constant $K$ that satisfies our unique curve.

Question

The solution of the differential equation $(x^2 + y^2) dx - 5xy dy = 0, y(1) = 0$, is :

Options:

(A) $|x^2 - 2y^2|^6 = x$

(B) $|x^2 - 4y^2|^6 = x$

(C) $|x^2 - 4y^2|^5 = x^2$

(D) $|x^2 - 2y^2|^5 = x^2$

Sigma Percentile · Accepted Answer

### The Art of the Homogeneous Transformation

Welcome, fellow JEE aspirant. Today, we are not just solving a differential equation; we are embarking on a journey of pattern recognition.

When you first look at the equation $(x^2 + y^2) dx - 5xy dy = 0$, it might seem like a chaotic jumble of variables. But to the trained eye, this is a beautiful, structured puzzle waiting to be solved.

### Phase 1: The Diagnosis

Before we touch a single piece of algebra, we must diagnose the problem. Look at the terms: $x^2$, $y^2$, and $5xy$.

Notice something? Every single term has a total degree of 2. This is the hallmark of a **homogeneous differential equation**.

In the world of JEE, recognizing this is half the battle. It tells us that the equation has a specific symmetry, and that symmetry is our golden ticket to a solution.

We start by rearranging the equation to isolate the derivative, $\frac{dy}{dx}$:

$$5xy \frac{dy}{dx} = x^2 + y^2 \implies \frac{dy}{dx} = \frac{x^2 + y^2}{5xy}$$

### Phase 2: The Transformation

Now, we introduce our most powerful tool: the substitution $y = vx$. Why do we do this? Because it transforms a non-separable equation into a separable one.

If $y = vx$, then by the product rule, the derivative is:

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Let us substitute this into our equation. This is where you must be meticulous. Do not rush. Replace every $y$ with $vx$:

$$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{5x(vx)}$$

Watch the magic happen. The $x^2$ terms in the numerator and denominator are identical. They cancel out, leaving us with a clean, manageable expression:

$$v + x \frac{dv}{dx} = \frac{1 + v^2}{5v}$$

### Phase 3: The Calculus Battle

We are now in the home stretch. We need to isolate the variables. Move the $v$ to the right side:

$$x \frac{dv}{dx} = \frac{1 + v^2}{5v} - v = \frac{1 + v^2 - 5v^2}{5v} = \frac{1 - 4v^2}{5v}$$

Now, separate the variables. Bring all the $v$ terms to the left and all the $x$ terms to the right:

$$\frac{5v}{1 - 4v^2} dv = \frac{dx}{x}$$

This is the moment of truth. Integrating the right side is trivial: $\ln |x| + C$. But the left side requires a steady hand.

We recognize that the derivative of the denominator $(1 - 4v^2)$ is $-8v$. We have $5v$ in the numerator. By adjusting the constants, we get:

$$-\frac{5}{8} \ln |1 - 4v^2| = \ln |x| + C$$

### Phase 4: The Final Polish

We are almost at the finish line. We need to clean up these logarithms. Multiply by $-8$ and use the properties of logs to bring the coefficients inside as exponents:

$$5 \ln |1 - 4v^2| = -8 \ln |x| + C'$$

Exponentiating both sides allows us to remove the logs, leading us to:

$$|1 - 4v^2|^5 = K x^{-8}$$

Substituting back $v = \frac{y}{x}$, we get:

$$\left|1 - 4\frac{y^2}{x^2}\right|^5 = K x^{-8} \implies \left|\frac{x^2 - 4y^2}{x^2}\right|^5 = K x^{-8}$$

Simplifying this, we arrive at the general solution:

$$|x^2 - 4y^2|^5 = K x^2$$

Finally, we apply our initial condition $y(1) = 0$. Plugging these values in, we find $K = 1$.

Our final, elegant solution is:

$$\mathbf{|x^2 - 4y^2|^5 = x^2}$$

This matches our target option perfectly. Remember, the math is not just about the final answer; it is about the journey of transformation. You have successfully navigated the complexity of a homogeneous differential equation.

Animated Solution for Mathematics - Differential Equations: The solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange to $\frac{d y}{d x}$ Form

Identify Homogeneity

Substitution Strategy

Apply Substitution

Simplify the Expression

Isolate $x \frac{d v}{d x}$

Separate the Variables

Integrate Both Sides

Apply Logarithmic Properties

Remove Logarithms

Back Substitution

Final Algebraic Simplification

Apply Initial Condition

Final Solution

The Sigma Insight: Homogeneous Differential Equations

The Art of the Homogeneous Transformation

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; } The solution of the differential equation (x2+y2)dx−5xydy=0,y(1)=0, is :