Question

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y}$, where $0\le x<\frac{\pi }{2}$ 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 $y^2=1-\frac{x}{\sec x+\tan x}$

$\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y}$
$2y\frac{dy}{dx}+y^2\sec x=\tan x$
Put $y^2=t\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}$
$\frac{dt}{dx}+t\sec x=\tan x$
$I.f=e^{\int \sec xdx}=e^{In(\sec x+\tan x)}=\sec x+\tan x$
$\frac{dt}{dx}(\sec x+\tan x)+t\sec x(\sec x+\tan x)=\tan x(\sec x+\tan x)$
$\int d\left(t\right(\sec x+\tan x\left)\right)=\tan x(\sec x+\tan x)dx$
$t(\sec x+\tan x)=\sec x+\tan x-x$
$t=1-\frac{x}{\sec x+\tan x},y^2=1-\frac{x}{\sec x+\tan x}$