Question

The general solutions of the differential equation $\sin 2x(\frac{dy}{dx}-\sqrt{tanx})-y=0$ is

• A. $y\sqrt{\tan x}=x+c^2$
• B. $y\sqrt{\cot x}=\tan x+c^2$
• C. $y\sqrt{\tan x}=\cot x+c^4$
• D. $y\sqrt{\cot x}=x+c$

Answer 1

The correct option is D $y\sqrt{\cot x}=x+c$

Given,

$sin2x(\frac{dy}{dx}-\sqrt{tanx})-y=0$

Divide above equation by sin2x

$\frac{dy}{dx}-\frac{y}{sin2x}=\sqrt{tanx}$

P=-$\frac{1}{sin2x}$ and Q=$\sqrt{tanx}$

I.f.=$e^{\int P.dx}$

=$e^{\int -\frac{1}{sin2x}dx}$

=$e^{\int -\frac{se{c}^{2}x}{2tanx}dx}$

Let tanx=t

$se{c}^{2}xdx=dt$

$\therefore I.F.=$e^{\int -\frac{1}{2t}dt}$

=$e^{-\frac{1}{2}logt}$

=$t^{\frac{-1}{2}}$

= $\frac{1}{\sqrt{tanx}}$

Complete solution

$y\times \frac{1}{\sqrt{tanx}}=\int dx+c$

$\frac{y}{\sqrt{tanx}}=x+c$