Question

The solution of $sin2x$ $\left(\frac{dy}{dx}\right)-y=\tan x$ is:

• A. $x-ysinx=c$
• B. $xytanx=c$
• C. $y=tanx+c$
• D. $y=tanx+c$ $\sqrt{tanx}$

Answer 1

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

$\sin 2x\frac{dy}{dx}-y=\tan x\Rightarrow \frac{dy}{dx}-y\csc 2x=\csc 2x\tan x$
Substituting $t=e^{-\int \csc 2xdx}=\sqrt{\frac{\cos x}{\sin x}}$
We get
$t\frac{dy}{dx}-yt\csc 2x=t\csc 2x\tan x\Rightarrow \sqrt{\frac{\cos x}{\sin x}}\frac{dy}{dx}-y\sqrt{\frac{\cos x}{\sin x}}\csc 2x=\sqrt{\frac{\sin x}{\cos x}}\csc 2x$
Now substituting $-\sqrt{\frac{\cos x}{\sin x}}\csc 2x=\frac{d}{dx}\left(\sqrt{\frac{\cos x}{\sin x}}\right)$, we get
$\sqrt{\frac{\cos x}{\sin x}}\frac{dy}{dx}-y\frac{d}{dx}\left(\sqrt{\frac{\cos x}{\sin x}}\right)\csc 2x=\sqrt{\frac{\sin x}{\cos x}}\csc 2x$
Using formula $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(\sqrt{\frac{\cos x}{\sin x}}y\right)=\sqrt{\frac{\sin x}{\cos x}}\csc 2x$
Integrating both sides
$\int \frac{d}{dx}\left(\sqrt{\frac{\cos x}{\sin x}}y\right)dx=\int \sqrt{\frac{\sin x}{\cos x}}\csc 2xdx\Rightarrow \sqrt{\frac{\cos x}{\sin x}}y=\sqrt{\frac{\sin x}{\cos x}}+c\Rightarrow y=\tan x+c\sqrt{\tan x}$