Question

The general solution of the differential equation, $\sin 2x(\frac{dy}{dx}-\sqrt{\tan x})-y=0$, 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. $y\sqrt{\cot x}=x+c$

Answer 1

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

$\sin 2x(y^{\prime }-\sqrt{\tan x})+y=0$

$y^{\prime }=\sqrt{\tan x}-\frac{y}{\sin 2x}$

$y^{\prime }+\frac{y}{\sin 2x}=\sqrt{\tan x}$

If $=e^{\int f\left(x\right)dx}=e^{\int \frac{dx}{\sin 2x}}=$

$=e^{\int \frac{{\sec }^{2}x}{2tanx}dx}=e^{\int \frac{dt}{2t}}$ where $t=\tan x$......(solve the above integration we get)

$=\sqrt{\cot x}$

$y(I.F.)=(I.F.)\int g\left(x\right)dx+c$

$y\sqrt{\cot x}=x+c$