Q. The solution of $2(y + 3) - xy\ \frac{dy}{dx} = 0$ with $y = - 2$, where $x = 1$, is
A) $y + 3 = x^{2}$
B) $x^{2}(y + 3) = 1$
C) $x^{4}(y + 3) = 1$
D) $x^{2}(y + 3)^{3} = e^{y + 2}$
Explanation
(d)
$\because\ 2(y + 3) - xy\ \frac{dy}{dx} = 0$
$\Rightarrow 2(y + 3) = xy\ \frac{dy}{dx}$
$\Rightarrow \ \int_{}^{}{\frac{2}{x}\ dx = \int_{}^{}{\frac{y}{y + 3}dx}}$
$\Rightarrow 2\log{x = y - 3\log{(y + 0) + c}}$
Put $x = 1$ and $y = - 2$
$\Rightarrow 2 = c$
$\therefore x^{2}(y + 3)^{3} = e^{y + 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.