A curve passes through the point (1, π/6). Let the slope of the curve at each point (x, y) be y/x +

Q: Why do we use $y = vx$ instead of $x = vy$?

Great question! We use $y = vx$ because our differential equation is defined as $\frac{dy}{dx}$. If the problem had given us $\frac{dx}{dy}$, we would have used $x = vy$ to make the integration easier. It is all about choosing the path of least resistance to isolate the derivative.

Q: Wait, why did the $v$ terms cancel out? Is that a coincidence?

It is not a coincidence at all! That is the defining characteristic of a homogeneous differential equation. The substitution is specifically designed to isolate the derivative term, allowing the variables to be separated. If they did not cancel, we would know we made a mistake in our substitution or identification.

Q: Why do we need the point $(1, \pi/6)$?

When we integrate, we get an indefinite integral, which introduces an arbitrary constant $C$. This constant represents a family of curves. The point $(1, \pi/6)$ acts as a 'pin' that fixes our curve to a specific location, allowing us to determine the unique value of $C$ for this specific curve.

Question

A curve passes through the point $(1, \pi/6)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + \sec(\frac{y}{x}), x > 0$. Then the equation of the curve is

Options:

(A) $\sin(\frac{y}{x}) = \ln x + \frac{1}{2}$

(B) $\csc(\frac{y}{x}) = \ln x + 2$

(C) $\sec(\frac{2y}{x}) = \ln x + 2$

(D) $\cos(\frac{2y}{x}) = \ln x + \frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a blank coordinate plane. You are tasked with finding a curve that weaves through this space, governed by the slope rule:

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

At first glance, this might look intimidating. However, observe the right-hand side: every term is a function of the ratio $\frac{y}{x}$.

This is not random; it is a structural signature. This is a **Homogeneous Differential Equation**. Recognizing this is your first victory.

### The Power of Substitution

Now that we have identified the structure, we use the standard substitution $y = vx$. This is a classic move in the JEE toolkit.

By setting $y = vx$, we change our perspective from the coordinates $(x, y)$ to the variable $x$ and the slope-ratio $v$. We must also transform the derivative $\frac{dy}{dx}$ using the product rule:

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

This expression acts as the bridge that connects our old coordinate system to the new one.

### The Algebraic Dance

Now, let us perform the substitution. We replace $\frac{dy}{dx}$ with $v + x\frac{dv}{dx}$ and $\frac{y}{x}$ with $v$:

$$v + x\frac{dv}{dx} = v + \sec(v)$$

Watch closely as the magic happens. The $v$ on the left and the $v$ on the right are identical and cancel out perfectly. We are left with:

$$x\frac{dv}{dx} = \sec(v)$$

This is the beauty of the homogeneous method; it strips away the complexity, leaving us with a simple, **separable differential equation**.

### The Path to Integration

We are now in the home stretch. We separate the variables:

$$\frac{dv}{\sec(v)} = \frac{dx}{x}$$

Recall your trigonometry: $\frac{1}{\sec(v)}$ is simply $\cos(v)$. Our equation becomes:

$$\int \cos(v) \, dv = \int \frac{1}{x} \, dx$$

Integrating both sides yields:

$$\sin(v) = \ln|x| + C$$

We must return to our original variables by substituting $v = \frac{y}{x}$ back into the equation:

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

### The Final Pin

We have a family of curves, but we need the specific one that passes through the point $(1, \pi/6)$. We substitute $x = 1$ and $y = \frac{\pi}{6}$ into our general solution:

$$\sin\left(\frac{\pi/6}{1}\right) = \ln(1) + C$$

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\ln(1) = 0$, we find that $C = \frac{1}{2}$.

The final equation of the curve is:

$$\mathbf{\sin\left(\frac{y}{x}\right) = \ln x + \frac{1}{2}}$$

You have successfully navigated the complexity, identified the structure, and solved the mystery. This is the essence of JEE Advanced mathematics: not just solving, but understanding the underlying architecture of the problem.

Animated Solution for Mathematics - Differential Equations: A curve passes through the point $(1, π /6)$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + sec (\frac{y}{x}), x > 0$ . Then the equation of the curve is

Select Answer:

Visualized Solution

Initial Condition

Setting up the Differential Equation

Identifying the Type of ODE

The Standard Substitution

Differentiating the Substitution

Substituting Back into the ODE

Simplifying the Equation

Separating the Variables

Preparing for Integration

Integrating Both Sides

Re-substituting $v = \frac{y}{x}$

Applying the Initial Condition

Calculating the Constant $C$

Final Equation of the Curve

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Power of Substitution

The Algebraic Dance

The Path to Integration

The Final Pin

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