Question

The general solution of the differential equation $\frac{dy}{dx} = 2^{y - x}$ is

• A. $2^x - 2^y = c$
• B. $\frac{1}{2^x} - \frac{1}{2^y} = c$
• C. $\frac{1}{2^x} + \frac{1}{2^y} = c$
• D. $2^x + 2^y = c$

Hint:

Pay close attention to signs during variable separation! Because both exponential terms move to denominators or change signs during integration ($2^{-y} \rightarrow -2^{-y}$), the final balance must feature a subtraction sign between the two variable blocks. This allows you to immediately eliminate options (C) and (D).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the general algebraic solution for a first-order variable-separable differential equation where the variables are in the exponents of base 2.

Step 2: Key Formula or Approach:
1. Use exponent rules to separate the right-hand side: $2^{y-x} = \frac{2^y}{2^x}$.
2. Separate variables by moving all $y$ terms to the left side and all $x$ terms to the right side: $$\frac{1}{2^y} dy = \frac{1}{2^x} dx \implies 2^{-y} dy = 2^{-x} dx$$ 3. Integrate both sides using the standard exponential integration rule: $$\int a^u du = \frac{a^u}{\log_e a}$$

Step 3: Detailed Explanation:
4. Separate variables in the differential equation: $$\frac{dy}{dx} = \frac{2^y}{2^x} \implies \frac{1}{2^y} dy = \frac{1}{2^x} dx$$ $$2^{-y} dy = 2^{-x} dx$$ 5. Integrate both sides step-by-step: $$\int 2^{-y} dy = \int 2^{-x} dx$$ $$\frac{2^{-y}}{-\log_e 2} = \frac{2^{-x}}{-\log_e 2} + c_1$$ 6. Multiply the entire equation by $-\log_e 2$ to eliminate the denominators: $$2^{-y} = 2^{-x} - c_1 \log_e 2$$ Rearrange the terms to group the variables on one side: $$2^{-x} - 2^{-y} = c_1 \log_e 2$$ 7. Rewrite negative exponents as fractions and define a new consolidated constant $c = c_1 \log_e 2$: $$\frac{1}{2^x} - \frac{1}{2^y} = c$$

Step 4: Final Answer:
The general solution of the differential equation matches option (B).