If dy/dx + (2^(x-y)(2ʸ - 1)) / (2ˣ - 1) = 0, x, y 0, y(1) = 1, then y(2) is equal to : (A) 2 + log

Q: Why can't we just use the integrating factor method here?

That is a fantastic question! The integrating factor method is specifically designed for *linear* differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$. If you look closely at our equation, the $2^y$ term makes it non-linear. Always check for linearity before jumping into the integrating factor formula; otherwise, you will be chasing a solution that doesn't exist!

Q: Why does the constant $C$ turn into $K$ after exponentiation?

When we exponentiate both sides to remove the logarithms, we get $e^{\ln|A| + \ln|B|} = e^C$. Since $e^C$ is just another constant, we rename it $K$ to keep our algebra clean. It is a standard practice to simplify the expression and make it easier to solve for the initial condition.

Q: Why do we divide by $\ln 2$ during integration?

Remember the chain rule! The derivative of $2^x$ is $2^x \ln 2$. When we integrate $\frac{2^x}{2^x - 1} dx$, we are essentially looking for the integral of $\frac{1}{u} du$ where $u = 2^x - 1$. Since $du = 2^x \ln 2 dx$, we have $dx = \frac{du}{2^x \ln 2}$. The $\ln 2$ is a constant factor that must be accounted for to balance the equation.

Question

If $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0, x, y > 0, y(1) = 1$, then $y(2)$ is equal to :

Options:

(A) $2 + \log_2 3$

(B) $2 + \log_2 2$

(C) $2 - \log_2 3$

(D) $2 - \log_2 3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponents.

When you see an equation like $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$, it is natural to feel a moment of hesitation. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance.

Let us peel back that mask together.

### The Art of Decoupling

The first step in any battle is to organize your forces. We have a differential equation, and our goal is to isolate the derivative. By moving the fraction to the right-hand side, we get:

$$\frac{dy}{dx} = -\frac{2^{x-y}(2^y - 1)}{2^x - 1}$$

Now, look at that term $2^{x-y}$. It is the anchor dragging us down. But we have a powerful tool in our arsenal: the laws of exponents.

We know that $a^{m-n} = \frac{a^m}{a^n}$. Applying this, we transform the equation into:

$$\frac{dy}{dx} = -\frac{2^x \cdot (2^y - 1)}{2^y \cdot (2^x - 1)}$$

Suddenly, the fog clears. We have successfully separated the $x$ terms from the $y$ terms. This is the **Variable Separable** method in its purest form.

By cross-multiplying, we bring all the $y$ terms to the left and all the $x$ terms to the right:

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

### The Integration Dance

Now that we have our variables in their respective corners, we integrate. This is where many students stumble, but you are not going to be one of them. We are looking at integrals of the form $\int \frac{f'(x)}{f(x)} dx$.

Recall that the derivative of $2^x$ is $2^x \ln 2$. Our numerator is $2^x$, which is missing that $\ln 2$ factor. So, we introduce it by multiplying and dividing by $\ln 2$.

The integral becomes:

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

See how the $\ln 2$ appears on both sides? We can multiply the entire equation by $\ln 2$ to clear the denominators. The constant $C$ multiplied by $\ln 2$ is still just a constant, which we can call $C'$.

Now we have:

$$\ln |2^y - 1| + \ln |2^x - 1| = C'$$

### The Logarithmic Bridge

We are nearing the finish line. Using the property $\ln A + \ln B = \ln(AB)$, we combine the left side into a single logarithm:

$$\ln |(2^y - 1)(2^x - 1)| = C'$$

To liberate our variables from the logarithm, we exponentiate both sides. $e^{C'}$ is just another constant, which we will call $K$.

Thus, we arrive at our general solution:

$$(2^y - 1)(2^x - 1) = K$$

### The Final Reveal

We are given the initial condition $y(1) = 1$. This means when $x = 1$, $y = 1$. Let us plug these values in to find the identity of $K$:

$$(2^1 - 1)(2^1 - 1) = K \Rightarrow (1)(1) = K \Rightarrow K = 1$$

Our particular solution is now locked in: $(2^y - 1)(2^x - 1) = 1$. The final question asks for $y(2)$, so we set $x = 2$:

$$(2^y - 1)(2^2 - 1) = 1$$

$$(2^y - 1)(3) = 1$$

Solving for $2^y$:

$$2^y - 1 = \frac{1}{3} \Rightarrow 2^y = 1 + \frac{1}{3} = \frac{4}{3}$$

Finally, we take the logarithm base 2 of both sides:

$$y = \log_2 \left(\frac{4}{3}\right) = \log_2 4 - \log_2 3$$

Since $\log_2 4 = 2$, we get our final, elegant answer:

**$$y(2) = 2 - \log_2 3$$**

And there you have it. What seemed like a daunting differential equation was simply a sequence of logical steps, each building upon the last. Keep this clarity of mind, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Apply Laws of Exponents

Separate the Variables

Integrate Both Sides

Apply Substitution Method

Simplify Using Log Properties

Formulate the General Solution

Find the Constant of Integration

Substitute $x = 2$ to find $y (2)$

Solve for $y$

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Decoupling

The Integration Dance

The Logarithmic Bridge

The Final Reveal

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