Key Concepts and Formulas

• Variable Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = \frac{f(x)}{g(y)}$ can be solved by separating variables and integrating.
• Integration Formulas: $\int \frac{1}{x} dx = \ln|x| + C$, $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$
• Logarithm Properties: $\log_a b = \frac{\ln b}{\ln a}$, $a^{\log_a x} = x$, $\log_a x + \log_a y = \log_a (xy)$, $\log_a x - \log_a y = \log_a (\frac{x}{y})$

Step-by-Step Solution

Step 1: Separate the variables

We are given the differential equation: $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$ Our goal is to separate the variables $x$ and $y$. First, subtract the second term from both sides: $\frac{dy}{dx} = - \frac{2^{x-y}(2^y - 1)}{2^x - 1}$ Rewrite $2^{x-y}$ as $\frac{2^x}{2^y}$: $\frac{dy}{dx} = - \frac{2^x (2^y - 1)}{2^y (2^x - 1)}$ Now, separate the variables: $\frac{2^y}{2^y - 1} dy = - \frac{2^x}{2^x - 1} dx$

Step 2: Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $\int \frac{2^y}{2^y - 1} dy = - \int \frac{2^x}{2^x - 1} dx$ Let's evaluate the left-hand side integral. Let $u = 2^y - 1$, then $du = 2^y \ln 2 \, dy$, so $2^y dy = \frac{du}{\ln 2}$. Therefore, $\int \frac{2^y}{2^y - 1} dy = \int \frac{1}{u} \frac{du}{\ln 2} = \frac{1}{\ln 2} \int \frac{1}{u} du = \frac{1}{\ln 2} \ln |u| + C_1 = \frac{1}{\ln 2} \ln |2^y - 1| + C_1$ Similarly, for the right-hand side integral, let $v = 2^x - 1$, then $dv = 2^x \ln 2 \, dx$, so $2^x dx = \frac{dv}{\ln 2}$. Therefore, $- \int \frac{2^x}{2^x - 1} dx = - \int \frac{1}{v} \frac{dv}{\ln 2} = - \frac{1}{\ln 2} \int \frac{1}{v} dv = - \frac{1}{\ln 2} \ln |v| + C_2 = - \frac{1}{\ln 2} \ln |2^x - 1| + C_2$ Combining these results, we have: $\frac{1}{\ln 2} \ln |2^y - 1| = - \frac{1}{\ln 2} \ln |2^x - 1| + C$ where $C = C_2 - C_1$. Multiply both sides by $\ln 2$: $\ln |2^y - 1| = - \ln |2^x - 1| + C \ln 2$ Let $K = C \ln 2$, then $\ln |2^y - 1| = - \ln |2^x - 1| + K$ $\ln |2^y - 1| + \ln |2^x - 1| = K$ $\ln |(2^y - 1)(2^x - 1)| = K$ $(2^y - 1)(2^x - 1) = e^K = A$ where $A$ is a constant.

Step 3: Apply the initial condition

We are given that $y(1) = 1$. Substitute $x = 1$ and $y = 1$ into the equation: $(2^1 - 1)(2^1 - 1) = A$ $(2 - 1)(2 - 1) = A$ $(1)(1) = A$ $A = 1$ So, the equation becomes: $(2^y - 1)(2^x - 1) = 1$

Step 4: Find y(2)

We want to find $y(2)$, so we substitute $x = 2$ into the equation: $(2^y - 1)(2^2 - 1) = 1$ $(2^y - 1)(4 - 1) = 1$ $(2^y - 1)(3) = 1$ $2^y - 1 = \frac{1}{3}$ $2^y = 1 + \frac{1}{3} = \frac{4}{3}$ $y = \log_2 \left( \frac{4}{3} \right)$ $y = \log_2 4 - \log_2 3$ $y = 2 - \log_2 3$ This is $y(2)$.

Step 5: Correct the Solution.

Upon re-examination, there was an error in the solution above. Let's correct it.

From Step 2, we have: $\frac{1}{\ln 2} \ln |2^y - 1| = - \frac{1}{\ln 2} \ln |2^x - 1| + C$ Multiplying by $\ln 2$ gives: $\ln |2^y - 1| = - \ln |2^x - 1| + C \ln 2$ Using the initial condition $y(1) = 1$, we get: $\ln |2^1 - 1| = - \ln |2^1 - 1| + C \ln 2$ $\ln |1| = - \ln |1| + C \ln 2$ $0 = - 0 + C \ln 2$ $C = 0$ So, the equation becomes: $\ln |2^y - 1| = - \ln |2^x - 1|$ $\ln |2^y - 1| + \ln |2^x - 1| = 0$ $\ln |(2^y - 1)(2^x - 1)| = 0$ $(2^y - 1)(2^x - 1) = e^0 = 1$ Now, let's find $y(2)$. Substitute $x = 2$: $(2^y - 1)(2^2 - 1) = 1$ $(2^y - 1)(4 - 1) = 1$ $(2^y - 1)(3) = 1$ $2^y - 1 = \frac{1}{3}$ $2^y = \frac{4}{3}$ $y = \log_2 \frac{4}{3} = \log_2 4 - \log_2 3 = 2 - \log_2 3$

Step 6: Re-Examine the Given Answer and Correct the Approach

The correct answer is given as $2 + \log_2 3$. The current derived answer is $2 - \log_2 3$. There must be an error.

Let's go back to the original equation and carefully re-examine the steps.

$\frac{dy}{dx} + \frac{2^{x-y} (2^y - 1)}{2^x - 1} = 0$ $\frac{dy}{dx} = - \frac{2^x (2^y - 1)}{2^y (2^x - 1)}$ $\frac{2^y - 1}{2^y} dy = - \frac{2^x}{2^x - 1} dx$ (ERROR HERE. It should be $\frac{2^y}{2^y - 1} dy$ as before).

Let's start again: $\frac{dy}{dx} = - \frac{2^{x-y}(2^y - 1)}{2^x - 1}$ $\frac{dy}{dx} = - \frac{2^x}{2^y} \frac{(2^y - 1)}{2^x - 1}$ $\frac{2^y}{2^y - 1} dy = - \frac{2^x}{2^x - 1} dx$ Integrating both sides: $\int \frac{2^y}{2^y - 1} dy = - \int \frac{2^x}{2^x - 1} dx$ Let $u = 2^y - 1$, $du = 2^y \ln 2 dy$, so $\frac{1}{\ln 2} \int \frac{du}{u} = \frac{1}{\ln 2} \ln |2^y - 1|$. Let $v = 2^x - 1$, $dv = 2^x \ln 2 dx$, so $- \frac{1}{\ln 2} \int \frac{dv}{v} = - \frac{1}{\ln 2} \ln |2^x - 1|$. So, $\frac{1}{\ln 2} \ln |2^y - 1| = - \frac{1}{\ln 2} \ln |2^x - 1| + C$. $\ln |2^y - 1| = - \ln |2^x - 1| + C \ln 2$. $(2^y - 1)(2^x - 1) = e^{C \ln 2} = A$. Given $y(1) = 1$, $(2^1 - 1)(2^1 - 1) = 1$, so $A = 1$. $(2^y - 1)(2^x - 1) = 1$. $x = 2$, $(2^y - 1)(2^2 - 1) = 1$, so $(2^y - 1)(3) = 1$, $2^y - 1 = \frac{1}{3}$, $2^y = \frac{4}{3}$, $y = \log_2 \frac{4}{3} = \log_2 4 - \log_2 3 = 2 - \log_2 3$.

Still not the correct answer! Let's try a different approach.

Going back to $\frac{dy}{dx} + \frac{2^{x-y} (2^y - 1)}{2^x - 1} = 0$: $\frac{dy}{dx} = - \frac{2^x}{2^y} \frac{(2^y - 1)}{(2^x - 1)} = \frac{2^x - 2^{x-y}}{2^x - 1}$. If $y = x$, then $\frac{dy}{dx} = 1$. However, $\frac{2^x - 1}{2^x - 1} = 1$, so $\frac{dy}{dx} = -1$. This is incorrect.

Let $u = 2^x$ and $v = 2^y$. Then $\frac{dv}{dx} = \frac{dv}{dy} \frac{dy}{dx} = v \ln 2 \frac{dy}{dx}$. So, $\frac{dy}{dx} = \frac{1}{v \ln 2} \frac{dv}{dx}$. Then we have: $\frac{1}{v \ln 2} \frac{dv}{dx} + \frac{u/v (v - 1)}{u - 1} = 0$. $\frac{dv}{dx} = - \frac{u (v - 1) v \ln 2}{v (u - 1)} = - \frac{u (v - 1) \ln 2}{u - 1}$. $\frac{dv}{v - 1} = - \frac{u \ln 2}{u - 1} dx$. $\int \frac{dv}{v - 1} = \ln |v - 1|$. $- \ln 2 \int \frac{u}{u - 1} dx = - \ln 2 \int \frac{u - 1 + 1}{u - 1} dx = - \ln 2 \int 1 + \frac{1}{u - 1} dx = - \ln 2 [x + \int \frac{1}{2^x - 1} dx]$. This looks very complicated.

Let's try the original approach again. $\frac{2^y}{2^y - 1} dy = - \frac{2^x}{2^x - 1} dx$. Integrating, $\frac{1}{\ln 2} \ln |2^y - 1| = - \frac{1}{\ln 2} \ln |2^x - 1| + C$. $y(1) = 1$, $\frac{1}{\ln 2} \ln |2^1 - 1| = - \frac{1}{\ln 2} \ln |2^1 - 1| + C$. $0 = - 0 + C$, so $C = 0$. Then $\ln |2^y - 1| = - \ln |2^x - 1|$. $(2^y - 1)(2^x - 1) = 1$. $x = 2$, $(2^y - 1)(3) = 1$, $2^y = \frac{4}{3}$, $y = \log_2 \frac{4}{3} = 2 - \log_2 3$.

There is STILL an error. Let's rethink this from the start.

Given $\frac{dy}{dx} + \frac{2^{x-y} (2^y - 1)}{2^x - 1} = 0$. $\frac{dy}{dx} = - \frac{2^x (2^y - 1)}{2^y (2^x - 1)}$. $\frac{2^y}{2^y - 1} dy = - \frac{2^x}{2^x - 1} dx$. $\int \frac{2^y}{2^y - 1} dy = - \int \frac{2^x}{2^x - 1} dx$. Let $I = \int \frac{2^y}{2^y - 1} dy = \int \frac{2^y - 1 + 1}{2^y - 1} dy = \int 1 + \frac{1}{2^y - 1} dy = y + \int \frac{1}{2^y - 1} dy = y + \int \frac{1}{2^y - 1} \frac{2^{-y}}{2^{-y}} dy = y + \int \frac{2^{-y}}{1 - 2^{-y}} dy$. Let $z = 1 - 2^{-y}$. $dz = 2^{-y} \ln 2 dy$. $I = y + \frac{1}{\ln 2} \int \frac{dz}{z} = y + \frac{1}{\ln 2} \ln |1 - 2^{-y}|$. So, $y + \frac{1}{\ln 2} \ln |1 - 2^{-y}| = - x - \frac{1}{\ln 2} \ln |1 - 2^{-x}| + C$. Given $y(1) = 1$, $1 + \frac{1}{\ln 2} \ln |1 - 2^{-1}| = -1 - \frac{1}{\ln 2} \ln |1 - 2^{-1}| + C$. $2 + \frac{2}{\ln 2} \ln |\frac{1}{2}| = C$. $2 - \frac{2 \ln 2}{\ln 2} = C$, so $C = 0$. $y + \frac{1}{\ln 2} \ln |1 - 2^{-y}| = -x - \frac{1}{\ln 2} \ln |1 - 2^{-x}|$. $x = 2$, $y + \frac{1}{\ln 2} \ln |1 - 2^{-y}| = -2 - \frac{1}{\ln 2} \ln |1 - 2^{-2}| = -2 - \frac{1}{\ln 2} \ln |\frac{3}{4}|$.

This is also quite complex. Let's go back to $(2^y - 1)(2^x - 1) = 1$ from the variable separable method. This seems promising. $(2^y - 1)(2^x - 1) = 1$. So $2^y - 1 = \frac{1}{2^x - 1}$. $2^y = \frac{1}{2^x - 1} + 1 = \frac{2^x}{2^x - 1}$. $y = \log_2 \frac{2^x}{2^x - 1} = \log_2 2^x - \log_2 (2^x - 1) = x - \log_2 (2^x - 1)$. $y(1) = 1$, $1 = 1 - \log_2 (2^1 - 1) = 1 - \log_2 1 = 1 - 0 = 1$. Now, $y(2) = 2 - \log_2 (2^2 - 1) = 2 - \log_2 3$.

It seems that there is an error in the provided correct answer. All calculations consistently lead to $y(2) = 2 - \log_2 3$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when separating variables and integrating.
• Integration Constants: Always remember to include the constant of integration after performing indefinite integration. Use the initial condition to solve for the constant.
• Algebraic Manipulation: Double-check your algebraic manipulations to avoid making mistakes when rearranging the equation.

Summary

We solved the given differential equation by separating variables, integrating both sides, and using the initial condition to determine the constant of integration. After careful calculation, we found that $y(2) = 2 - \log_2 3$. However, the provided answer key states that the correct answer is $2 + \log_2 3$. There may be an error in the provided correct answer.

Final Answer

The derived answer is $2 - \log_2 3$, which corresponds to option (D). However, the provided correct answer is A, which is $2 + \log_2 3$.

There is likely an error in the problem statement or the provided answer key. The final answer is \boxed{2 - \log_2 3}.