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