Question

The particular solution of \( e^{\frac{dy}{dx}} = 2x + 1 \) given that \( y = 1 \) when \( x = 0 \) is:

• A. \( y = \left( x + \frac{1}{2} \right) \log |2x + 1| - x + 1 \)
• B. \( y = (x + 1) \log |2x + 1| - x + 1 \)
• C. \( y = \left( x + \frac{1}{2} \right) \log |2x + 1| - \frac{1}{2} x + 1 \)
• D. \( y = \left( x - \frac{1}{2} \right) \log |2x + 1| - x - 1 \)

Hint:

When solving differential equations involving logarithms, make sure to properly handle integration constants and apply initial conditions to determine the particular solution.

Answer 1

The Correct Option is A

We are given the differential equation \( e^{\frac{dy}{dx}} = 2x + 1 \), and we need to find the particular solution for the given initial condition \( y = 1 \) when \( x = 0 \). First, we need to solve the differential equation. Begin by integrating both sides of the equation. \[ e^{\frac{dy}{dx}} = 2x + 1 \] Taking the natural logarithm of both sides: \[ \frac{dy}{dx} = \log(2x + 1) \] Now, integrate both sides with respect to \( x \): \[ y = \int \log(2x + 1) \, dx \] This integral can be solved using integration by parts. After solving, we get the general solution: \[ y = \left( x + \frac{1}{2} \right) \log |2x + 1| + C \] Now, using the initial condition \( y = 1 \) when \( x = 0 \), we substitute these values to find \( C \): \[ 1 = \left( 0 + \frac{1}{2} \right) \log |2(0) + 1| + C \] Simplifying: \[ 1 = \frac{1}{2} \log |1| + C \quad \Rightarrow \quad 1 = C \] Thus, the particular solution is: \[ y = \left( x + \frac{1}{2} \right) \log |2x + 1| - x + 1 \] Therefore, the correct answer is option (A)