Question

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

• A. \(\log_2 (2 + e)\)
• B. \(\log_2 (2e)\)
• C. \(\log_2 (1 + e^2)\)
• D. \(\log_2 (1 + e)\)

Hint:

When exponential terms are mixed in a differential equation, factoring out common terms usually reveals a variable-separable structure.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a first-order ordinary differential equation which can be solved using the variable separation method.
Step 2: Detailed Explanation:
Given:
\[ \frac{dy}{dx} = \frac{2^x \cdot 2^y - 2^x}{2^y} = \frac{2^x (2^y - 1)}{2^y} \]
Separate the variables:
\[ \frac{2^y}{2^y - 1} \, dy = 2^x \, dx \]
Integrate both sides:
\[ \int \frac{2^y}{2^y - 1} \, dy = \int 2^x \, dx \]
Let \(u = 2^y - 1 \implies du = 2^y \ln 2 \, dy\):
\[ \frac{1}{\ln 2} \int \frac{1}{u} \, du = \int 2^x \, dx \implies \frac{1}{\ln 2} \ln|2^y - 1| = \frac{2^x}{\ln 2} + C \]
Multiply by \(\ln 2\):
\[ \ln(2^y - 1) = 2^x + C' \]
Given boundary condition \(y(0) = 1\):
\[ \ln(2^1 - 1) = 2^0 + C' \implies \ln(1) = 1 + C' \implies 0 = 1 + C' \implies C' = -1 \]
The general solution is \(\ln(2^y - 1) = 2^x - 1\).
To find \(y(1)\), set \(x = 1\):
\[ \ln(2^y - 1) = 2^1 - 1 = 1 \implies 2^y - 1 = e^1 = e \]
\[ 2^y = 1 + e \implies y = \log_2 (1 + e) \]
Step 3: Final Answer:
The value of \(y(1)\) is \(\log_2 (1 + e)\).