Q. The solution of the differential equation $\left( e^{- 2\sqrt{x}} - \frac{y}{\sqrt{x}} \right)\frac{dy}{dx} = 1$ is given by
A) $ye^{2\sqrt{x}} = x + c$
B) $ye^{- 2\sqrt{x}} = \sqrt{x} + c$
C) $y = \sqrt{x}$
D) $y = 3\sqrt{x}$
Explanation
(a)
Given differential equation can be rewritten as
$\frac{dy}{dx} + \frac{y}{\sqrt{x}} = e^{- 2\sqrt{x}}$
Here, $P = \frac{1}{\sqrt{x}},\ Q = e^{- 2\sqrt{x}}$
$\therefore$ $IF = e^{\int_{}^{}{\frac{1}{\sqrt{x}}\ dx}} = e^{2\sqrt{x}}$
$\therefore$ Solution is
$ye^{2x} = \int_{}^{}e^{2\sqrt{x}}e^{- 2\sqrt{x}}\ dx = \int_{}^{}1\ dx$
$\Rightarrow$ $ye^{2\sqrt{x}} = x + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.