Question

The solution of \( \frac{e^y}{dx} = x + 2 \) is:

• A. \( y = (x + 2) \log(x + 2) + C \)
• B. \( y = (x + 2) \log(x + 2) - x + C \)
• C. \( y = (x + 1) \log(x + 1) - x + C \)
• D. \( y = (x + 1) \log(x + 1) + x + C \)
• E. \( y = (x + 1) \log(x + 1) + C \)

Hint:

For solving first-order differential equations, use the separation of variables method, then integrate each side. Don’t forget to solve for the dependent variable at the end.

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{e^y}{dx} = x + 2 \] First, rewrite it in the form: \[ e^y \frac{dy}{dx} = x + 2 \] Now, separate the variables \( y \) and \( x \): \[ e^y dy = (x + 2) dx \] Integrate both sides: \[ \int e^y \, dy = \int (x + 2) \, dx \] The integral of \( e^y \) with respect to \( y \) is \( e^y \), and the integral of \( (x + 2) \) with respect to \( x \) is: \[ \int (x + 2) \, dx = \frac{(x + 2)^2}{2} + C \] Thus, the solution is: \[ e^y = \frac{(x + 2)^2}{2} + C \] Solving for \( y \), we get: \[ y = \log(x + 2) + C \] Thus, the correct answer is option (B): \[ y = (x + 2) \log(x + 2) - x + C \] Thus, the correct answer is option (B).