Question

The solution of the differential equation \(\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y}\), where \(0\le x\lt\frac{\pi}{2}\), 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. \(y=1+\frac{x}{\sec x+\tan x}\)

Hint:

Multiply by \(2y\) and put \(v=y^2\) to get a linear equation in \(v\); use the integrating factor \(\sec x+\tan x\).

Answer 1

The Correct Option is C

Multiply the equation by \(2y\): \(2y\frac{dy}{dx}+y^2\sec x=\tan x\).
Let \(v=y^2\), so \(\frac{dv}{dx}+v\sec x=\tan x\).
Integrating factor \(=e^{\int\sec x\,dx}=\sec x+\tan x\).
\(v(\sec x+\tan x)=\int\tan x(\sec x+\tan x)\,dx=\int(\sec x\tan x+\sec^2x-1)\,dx=\sec x+\tan x-x+c\).
Using \(y(0)=1\): \(1\cdot(1+0)=1+0-0+c\Rightarrow c=0\).
So \(y^2(\sec x+\tan x)=\sec x+\tan x-x\), i.e. \(y^2=1-\frac{x}{\sec x+\tan x}\).
Answer: (3) \(y^2=1-\frac{x}{\sec x+\tan x}\)