Question

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{\tan x}{2y}$, where $0 \leq x < \frac{\pi}{2}$, and $y (0) = 1$, is given by:

• A. $y = 1 - \frac{x}{\sec x + \tan x}$
• B. $y^2 = 1 - \frac{x}{\sec x + \tan x}$
• C. $y^2 = 1 - \frac{x}{\sec x + \tan x}$
• D. $y = 1 + \frac{x}{\sec x + \tan x}$

Answer 1

Answer: (C)

2y $\frac{d y}{d x}+y^{2} \sec x=\tan x$
put $y^{2}=t $
$\Rightarrow $ 2y $\frac{d y}{d x}=\frac{d t}{d x}$
$\frac{d t}{d x}+t \,\sec\, x=\tan \,x$
I.F. $=e^{[\sec\, x d x} = e ^{\ell n(\sec x+\tan x)}=\sec x+\tan x $
$ t. (\sec x+\tan x) =\int(\sec x+\tan x) \tan x d x$
$=\int \sec x \tan x d x+\int \tan ^{2} x d x $
$ y^{2}(\sec x+\tan x) =\sec x+\tan x-x+c $
$y(0)=1 $
$ \Rightarrow c=0 $
$ \Rightarrow y^{2}=1-\frac{x}{\sec x+\tan x} $