Q. If $\frac{dy}{dx} = 1 + x + y + xy$ and $y( - 1) = 0$, then the function $y$ is
A) $e^{\frac{(1 - x)^{2}}{2}}$
B) $e^{\frac{(1 + x)^{2}}{2}} - 1$
C) $\log_{e}{(1 + x)} - 1$
D) $(1 + x)$
Explanation
(b)
Given, $\frac{dy}{dx} = (1 + x)(1 + y)$
$\Rightarrow$ $\frac{1}{1 + y}dy = (1 + x)dx$
$\Rightarrow$ $\log{(1 + y)} = x + \frac{x^{2}}{2} + c$ [integrating]
At $y( - 1) = 0$
$\Rightarrow$ $c = \frac{1}{2}$
$\therefore$ $\log{(1 + y)} = \frac{x^{2} + 2x + 1}{2}$
$\Rightarrow$ $y = e^{\frac{(1 + x)^{2}}{2}} - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.