Q. The solution of the differential equation $e^{- x}(y + 1)dy + \left( \cos^{2}x - \sin{2x} \right)y\ dx = 0$ subjected to the condition that $y = 1$ when $x = 0$ is
A) $y + \log y + e^{x}\cos^{2}x = 2$
B) $\log{(y + 1)} + e^{x}\cos^{2}x = 1$
C) $y + \log y = e^{x}\cos^{2}x$
D) $(y + 1) + e^{x}\cos^{2}x = 2$
Explanation
(a)
Given, $\left( 1 + \frac{1}{y} \right)dy = - e^{x}\ \left( \cos^{2}x - \sin{2x} \right)dx$
On integrating both sides, we get
$y + \log y = - e^{x}\cos^{2}x + \int_{}^{}e^{x}\sin{2x}\ dx - \int_{}^{}e^{x}\sin{2x}\ dx + c$
$\Rightarrow$ $y + \log y = - e^{x}\cos^{2}x + c$
At x=0, y=1
$1 + 0 = - e^{0}\cos 0 + c \Rightarrow c = 2$
$\therefore$ Required solution is
$y + \log y = - e^{x}\cos^{2}x + 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.