Answer
$$
\frac{dy}{dx} \tan x =a+y , \quad y(\frac{\pi}{3})=a, \quad0\lt x \lt \frac{\pi}{2}
$$
the solution of the given differential equation is
$$
y=\frac{4a}{\sqrt 3} \sin x -a
$$
Work Step by Step
$$
\frac{dy}{dx} \tan x =a+y , \quad y(\frac{\pi}{3})=a, \quad0\lt x \lt \frac{\pi}{2}
$$
We write the equation in terms of differentials:
$$
\frac{dy}{a+y} =\frac{dx}{\tan x } , y+a \ne 0
$$
and integrate both sides:
$$
\int \frac{dy}{a+y}=\int\frac{dx}{\tan x }
$$
$ \Rightarrow$
$$
\ln |a+y|=\ln |\sin x|+C
$$
$\Rightarrow$
$$
|a+y|=e^{\ln |\sin x|+C}=e^{\ln |\sin x|} \cdot e^{C}=e^{C}|\sin x| \
$$
where $C $ is an arbitrary constant.
$$
a+y=\pm e^{C} \sin x=L \sin x
$$
where $L=\pm e^{C} $ is an arbitrary constant.
Now, to satisfy the condition $ y(\frac{\pi}{3})=a, $we have
$$
2a=L . \frac{\sqrt 3}{2} \Rightarrow L =\frac{4a}{\sqrt 3}
$$
Thus, the solution of the given differential equation is
$$
a+y=\frac{4a}{\sqrt 3} \sin x
$$
and so,
$$
y=\frac{4a}{\sqrt 3} \sin x -a
$$
