How do I find general solution for 

$$(1-\sqrt{3})\frac{dy}{dx} + y\sec x = y^{\sqrt3}\sec x$$

What I tried 

$$y^{-\sqrt3}(1-\sqrt{3})\frac{dy}{dx} + y^{1-\sqrt3}\sec x = \sec x$$

Then,
$u=y^{1-\sqrt3} \Rightarrow \frac{du}{dy}=(1-\sqrt3)y^{-\sqrt3}$
$\frac{du}{dx}+u\sec x=\sec x$

$$e^{\int \sec x \,dx}=e^{\ln |\sec x+\tan x|}=\sec x+\tan x$$

$$(\sec x+\tan x)\frac{du}{dx}+u(\sec^2x+\tan x\sec x)=sec^2x+\tan x\sec x$$

$$\frac{d{\bigl((\sec x+\tan x)u\bigr)}}{dx}=\sec^2x+\tan x\sec x$$

After integrating both sides - $\int dx$,
$$(\sec x+\tan x)u=\sec x+\tan x\\
u=1=y^{1-\sqrt3}\\
y=1$$

Where am I doing the mistake ?