Question

The solution of $e^{\frac{dy}{dx}} = x + 1, \, y(0) = 3$ is:

• A. $y - 2 = x\log x$
• B. $y - x - 3 = x\log x$
• C. $y - x - 3 = (x + 1)\log(x + 1)$
• D. $y + x - 3 = (x + 1)\log(x + 1)$

Answer 1

The Correct Option is D

The given differential equation is: \[ e^{\frac{dy}{dx}} = x + 1. \] Take the natural logarithm on both sides: \[ \frac{dy}{dx} = \ln(x + 1). \] Integrate both sides with respect to $x$: \[ y = \int \ln(x + 1) \, dx + C. \] Using integration by parts: \[ \int \ln(x + 1) \, dx = (x + 1)\ln(x + 1) - \int \frac{x + 1}{x + 1} \, dx = (x + 1)\ln(x + 1) - (x + 1). \] Thus: \[ y = (x + 1)\ln(x + 1) - (x + 1) + C. \] Apply the initial condition $y(0) = 3$: \[ 3 = (0 + 1)\ln(0 + 1) - (0 + 1) + C \implies C = 3. \] The solution becomes: \[ y = (x + 1)\ln(x + 1) - (x + 1) + 3 \implies y + x - 3 = (x + 1)\ln(x + 1). \]