What is the solution of the differential equation with y(0) = 0 ?
For the next two (02) items that follow : Consider the differential equation \( e^{x+y} \frac{dy}{dx} = e^{x-y} \)
- A. \( y = \ln(2x+1) \)
- B. \( y = \ln(2x-1) \)
- C. \( 2y = \ln(2x+1) \) ✓
- D. \( 2y = \ln(2x-1) \)
Correct Answer: C. \( 2y = \ln(2x+1) \)
Explanation
From the simplified equation \( e^{2y} dy = dx \), integrating both sides yields \( \frac{1}{2} e^{2y} = x + C \). Using y(0) = 0 gives \( C = 1/2 \). The equation becomes \( e^{2y} = 2x + 1 \), which simplifies to \( 2y = \ln(2x+1) \).
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