If sin(y/x) = log|x| + α/2 is the solution of the differential equation x cos(y/x) dy/dx = y cos(y/x

Q: Why did we divide by $x^2$? It feels like a random guess.

It is not random! It is pattern matching. We saw $x \frac{dy}{dx} - y$ and knew the quotient rule requires $x^2$ in the denominator. It is like finding the missing piece of a puzzle. When you see $x \frac{dy}{dx} - y$, your brain should immediately scream 'quotient rule'!

Q: Can I use $y=vx$ substitution instead?

Absolutely! That is the standard method for homogeneous equations. However, recognizing the exact differential is the 'pro' move. It saves time and reduces calculation errors. Both methods lead to the same result, but the exact differential method is more elegant.

Q: What happens to the absolute value in $\log_e |x|$?

The absolute value is there to ensure the domain of the logarithm is valid for negative $x$. In many JEE problems, we assume $x > 0$ for simplicity, but keeping the modulus is mathematically rigorous. It does not change the integration result for positive $x$.

Question

If $\sin(\frac{y}{x}) = \log_e |x| + \frac{\alpha}{2}$ is the solution of the differential equation $x \cos(\frac{y}{x}) \frac{dy}{dx} = y \cos(\frac{y}{x}) + x$ and $y(1) = \frac{\pi}{3}$, then $\alpha^2$ is equal to

Options:

(A) 3

(B) 12

(C) 4

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:

$$x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x$$

The presence of the term $\frac{y}{x}$ inside the cosine function is a clear indicator of a homogeneous structure. Rather than using standard substitution, we will look for a more elegant path.

### Phase 1: The Algebraic Surgery

First, we organize the terms by bringing them to one side:

$$x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} - y \cos\left(\frac{y}{x}\right) = x$$

Factoring out the common term $\cos\left(\frac{y}{x}\right)$, we obtain:

$$\cos\left(\frac{y}{x}\right) \left[ x \frac{dy}{dx} - y \right] = x$$

The expression inside the brackets, $\left[ x \frac{dy}{dx} - y \right]$, is the numerator of the quotient rule for the derivative of $\frac{y}{x}$.

### Phase 2: The Elegant Transformation

Recall that the derivative of $\frac{y}{x}$ with respect to $x$ is given by:

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

To utilize this, we divide both sides of our equation by $x^2$:

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

This simplifies to:

$$\cos\left(\frac{y}{x}\right) \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x}$$

By the chain rule, the left side is the derivative of $\sin\left(\frac{y}{x}\right)$ with respect to $x$. Thus, we have:

$$\frac{d}{dx} \left[ \sin\left(\frac{y}{x}\right) \right] = \frac{1}{x}$$

### Phase 3: The Final Integration

Integrating both sides with respect to $x$, we arrive at the general solution:

$$\sin\left(\frac{y}{x}\right) = \ln|x| + C$$

We apply the initial condition $y(1) = \frac{\pi}{3}$. Substituting $x=1$ and $y=\frac{\pi}{3}$:

$$\sin\left(\frac{\pi}{3}\right) = \ln|1| + C$$

Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\ln(1) = 0$, we find that **$C = \frac{\sqrt{3}}{2}$**.

### Phase 4: The Victory Lap

The specific solution is:

$$\sin\left(\frac{y}{x}\right) = \ln|x| + \frac{\sqrt{3}}{2}$$

Comparing this to the form $\sin\left(\frac{y}{x}\right) = \ln|x| + \frac{\alpha}{2}$, we identify $\frac{\alpha}{2} = \frac{\sqrt{3}}{2}$, which implies $\alpha = \sqrt{3}$.

Therefore, the final value is:

$$\alpha^2 = (\sqrt{3})^2 = \mathbf{3}$$

Animated Solution for Mathematics - Differential Equations: If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Rearranging the Terms

Factoring Out $cos (\frac{y}{x})$

Dividing by $x^{2}$

Recognizing the Quotient Rule

Identifying the Total Derivative

Integrating Both Sides

Applying Initial Conditions

Calculating the Constant $C$

Final Form of the Solution

Finding $α$ and $α^{2}$

Summary and Key Takeaway

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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