Q. The solution of $e^{\frac{dy}{dx}} = (x + 1),\ y(0) = 3$ is
A) $y = x\ \log{x - x} + 2$
B) $y = (x + 1)\log{|x + 1|} - x + 3$
C) $y = (x + 1)\log{|x + 1|} + x + 3$
D) $y = x\log x + x + 3$
Explanation
(b)
Given, $\frac{dy}{dx} = \log{(x + 1)}$
$\Rightarrow$ $dy = \log{(x + 1)}dx$
$\Rightarrow$ $\int_{}^{}{dy} = \int_{}^{}{\log{(x + 1)}dx}$
$\Rightarrow$ $y = (x + 1)\log{|x + 1| - x + c}$
$\therefore$ $x = 0,\ \ y = 3$
$\therefore$ $c = 3$
$\therefore$ $y = (x + 1)\log{|x + 1| - x + 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.