Question

The general solution of the differential equation, \(\sin2x\left(\frac{dy}{dx}-\sqrt{\tan x}\right)-y=0\), is:

• A. \(y\sqrt{\tan x}=x+c\)
• B. \(y\sqrt{\cot x}=\tan x+c\)
• C. \(y\sqrt{\tan x}=\cot x+c\)
• D. \(y\sqrt{\cot x}=x+c\)

Hint:

Rewrite as a linear equation \(\frac{dy}{dx}-\frac{y}{\sin2x}=\sqrt{\tan x}\) and use the integrating factor \(e^{-\int\csc2x\,dx}=\frac{1}{\sqrt{\tan x}}\).

Answer 1

The Correct Option is D

Divide by \(\sin2x\): \(\frac{dy}{dx}-\frac{y}{\sin2x}=\sqrt{\tan x}\).
\(\int\csc2x\,dx=\frac{1}{2}\ln\tan x\), so the integrating factor is \(e^{-\frac{1}{2}\ln\tan x}=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}\).
Then \(\frac{d}{dx}\left(y\sqrt{\cot x}\right)=\sqrt{\tan x}\cdot\sqrt{\cot x}=1\).
Integrating: \(y\sqrt{\cot x}=x+c\).
Answer: (4) \(y\sqrt{\cot x}=x+c\)