Question

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

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

Answer 1

The correct option is D $y^2=1-\frac{x}{\sec x+\tan x}$

$\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y}$
$\Rightarrow 2y\frac{dy}{dx}+y^2\sec x=\tan x$
Put $y^2=v\Rightarrow 2y\frac{dy}{dx}=\frac{dv}{dx}$
$\frac{dv}{dx}+v\sec x=\tan x$
$\text{I.F.}=e^{\int \sec xdx}$
$=e^{\ln (\sec x+\tan x)}=\sec x+\tan x$
The general solution is $v(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx$
$\Rightarrow y^2(\sec x+\tan x)=\sec x+\tan x-x+C$
$\text{Given,}y\left(0\right)=1\Rightarrow C=0$
$\therefore y^2=1-\frac{x}{\sec x+\tan x}$