Answer
$\displaystyle y=\frac{4a}{\sqrt{3}}\sin(x) - a = a(\frac{4}{\sqrt{3}}\sin(x) - 1)$
Work Step by Step
$\displaystyle y'\tan(x) = a + y$
Separation of variables:
$\displaystyle \frac{dy}{dx}\cdot\tan(x) = a + y$
$\displaystyle \frac{dy}{dx}= \frac{a+y}{\tan(x)}$
$\displaystyle \frac{1}{a+y}dy= \cot(x)dx$
Integrate with respect to each variable. For the right side, use $u$-substitution where $u = \sin(x)\quad$ so that $\quad du = \cos(x)dx$
$\displaystyle \int\frac{dy}{a+y}= \int\frac{\cos(x)dx}{\sin(x)}$
$\displaystyle \ln|a+y| = \int\frac{du}{u}$
$\displaystyle \ln|a+y| = \ln|u|+C$
$\displaystyle \ln|a+y| = \ln|\sin(x)|+C$
$\displaystyle e^{\ln|a+y|} = e^{\ln|\sin(x)|+C} = e^{\ln|\sin(x)|}\cdot e^C$
***Since $C$ has not taken on any value yet, we can rewrite $e^C$ as just $C$***
$\displaystyle a+y = C\sin(x)$
Plug in the initial condition and solve for $C$
$\displaystyle a+(a) = C\sin(\frac{\pi}{3})$
$\displaystyle 2a = C(\frac{\sqrt{3}}{2})$
$\displaystyle \frac{2a}{\frac{\sqrt{3}}{2}} = \frac{2a}{1} \cdot \frac{2}{\sqrt{3}}= C$
$\displaystyle C = \frac{4a}{\sqrt{3}}$
Plug the value for $C$ that we just found back into the equation and isolate $y$
$\displaystyle a+y=(\frac{4a}{\sqrt{3}})\sin(x)$
$\displaystyle y=\frac{4a}{\sqrt{3}}\sin(x) - a = a(\frac{4}{\sqrt{3}}\sin(x) - 1)$
