Let the solution curve y = y(x) of the differential equation, [x/√(x² - y²) + e^(y/x)] x dy/dx = x +

Q: Wait, why do we reject the negative root when integrating $\frac{1}{\sqrt{1-v^2}}$?

Great observation! In this specific problem, we are dealing with a curve passing through $(1,0)$ and $(2\alpha, \alpha)$ with $\alpha > 0$. The standard integral of $\frac{1}{\sqrt{1-v^2}}$ is $\sin^{-1}(v)$. We choose the principal value branch because the context of the problem (the points given) implies we are working within a domain where the function is well-defined and continuous.

Q: Why does the $Mv$ term cancel out? Is that always supposed to happen?

Yes! That is the 'magic' of the homogeneous substitution. When you substitute $y=vx$ into a homogeneous differential equation, the equation is specifically structured so that the $v$ terms that don't involve the derivative will always cancel out, allowing you to separate the variables $v$ and $x$. If they don't cancel, double-check your algebra—you might have missed a term!

Q: How do we handle the $\ln|x|$? Do we need to worry about the absolute value?

In this problem, we are given that the curve passes through $(1,0)$ and $(2\alpha, \alpha)$ with $\alpha > 0$. Since $x$ starts at $1$ and moves to $2\alpha$, $x$ is always positive. Therefore, $\ln|x|$ simply becomes $\ln(x)$. You can safely drop the absolute value bars.

Question

Let the solution curve $y = y(x)$ of the differential equation, $[\frac{x}{\sqrt{x^2 - y^2}} + e^{y/x}] x \frac{dy}{dx} = x + [\frac{x}{\sqrt{x^2 - y^2}} + e^{y/x}] y$ pass through the points $(1, 0)$ and $(2\alpha, \alpha), \alpha > 0$. Then $\alpha$ is equal to

Options:

(A) $\frac{1}{2} \exp(\frac{\pi}{6} + \sqrt{e} - 1)$

(B) $\frac{1}{2} \exp(\frac{\pi}{3} + \sqrt{e} - 1)$

(C) $\exp(\frac{\pi}{6} + \sqrt{e} + 1)$

(D) $2 \exp(\frac{\pi}{3} + \sqrt{e} - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, tangled knot of mathematical symbols. The differential equation

$$ \left[ \frac{x}{\sqrt{x^2 - y^2}} + e^{y/x} \right] x \frac{dy}{dx} = x + \left[ \frac{x}{\sqrt{x^2 - y^2}} + e^{y/x} \right] y $$

looks, at first glance, like a nightmare. It is a jumble of square roots, exponentials, and derivatives. But in the world of JEE Advanced, intimidation is often just a mask for elegance.

Let us peel back the layers of this problem together.

### The Homogeneous Insight

Whenever you see a differential equation where the variables $x$ and $y$ appear in ratios like $y/x$, your internal alarm bells should ring. This is the signature of a **homogeneous differential equation**.

Look closely at the term $\frac{x}{\sqrt{x^2 - y^2}}$. If we divide both the numerator and the denominator inside the square root by $x$, we get:

$$ \frac{1}{\sqrt{1 - (y/x)^2}} $$

Suddenly, the entire equation is written in terms of $y/x$. This is our golden ticket.

### The Substitution

To tame this beast, we employ our most trusted weapon: the substitution $y = vx$. This transforms our dependent variable $y$ into a new variable $v$, where $v = y/x$.

But we cannot just change $y$; we must also change the derivative. Using the product rule, we find that:

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

Let us define $M = \left[ \frac{1}{\sqrt{1 - v^2}} + e^v \right]$ to keep our workspace clean. Substituting these into our equation, we get:

$$ M \left( v + x \frac{dv}{dx} \right) = 1 + Mv $$

### The Magic Cancellation

Now, watch closely. If we expand the left side, we get $Mv + Mx \frac{dv}{dx} = 1 + Mv$.

Do you see it? The $Mv$ term appears on both sides of the equation. With a swift stroke of the pen, they cancel out, leaving us with the beautifully simple:

$$ Mx \frac{dv}{dx} = 1 $$

This is the moment where the complexity collapses, and the path forward becomes clear.

### Integration and the Anchor

With the variables separated, we have:

$$ \left[ \frac{1}{\sqrt{1 - v^2}} + e^v \right] dv = \frac{dx}{x} $$

Integrating both sides is now a standard procedure. The integral of $\frac{1}{\sqrt{1 - v^2}}$ is $\sin^{-1}(v)$, and the integral of $e^v$ is $e^v$. On the right, we have $\ln|x| + C$.

Substituting $v = y/x$ back in, we get:

$$ \sin^{-1}\left(\frac{y}{x}\right) + e^{y/x} = \ln|x| + C $$

We are given that the curve passes through $(1, 0)$. Plugging these values in, we find $\sin^{-1}(0) + e^0 = \ln(1) + C$, which simplifies to $0 + 1 = 0 + C$, so $C = 1$.

Our curve is now fully defined:

$$ \sin^{-1}\left(\frac{y}{x}\right) + e^{y/x} = \ln|x| + 1 $$

### The Final Reveal

Finally, we use the second point $(2\alpha, \alpha)$ to find $\alpha$. Substituting $x = 2\alpha$ and $y = \alpha$, we get:

$$ \sin^{-1}\left(\frac{1}{2}\right) + e^{1/2} = \ln(2\alpha) + 1 $$

Knowing that $\sin^{-1}(1/2) = \pi/6$ and $e^{1/2} = \sqrt{e}$, we have:

$$ \frac{\pi}{6} + \sqrt{e} = \ln(2\alpha) + 1 $$

Solving for $\alpha$, we get $\ln(2\alpha) = \frac{\pi}{6} + \sqrt{e} - 1$. This leads us to $2\alpha = \exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$, or:

$$ \alpha = \frac{1}{2} \exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right) $$

And there it is. What started as a terrifying jumble of terms has been reduced to a clean, elegant result. This is the beauty of mathematics—no matter how complex the problem, there is always a path to the truth if you stay calm and follow the logic.

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Simplifying to Homogeneous Form

Substitution: $y = v x$

Placing the Substitutions

Algebraic Cancellation

Variable Separation

Integrating Both Sides

Finding the Constant $C$

Evaluating $C$

Solving for $α$

Simplifying the Equation

Final Calculation

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Homogeneous Insight

The Substitution

The Magic Cancellation

Integration and the Anchor

The Final Reveal

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