## Calculus: Early Transcendentals 8th Edition

$$\frac{dy}{dx} \tan x =a+y , \quad y(\frac{\pi}{3})=a, \quad0\lt x \lt \frac{\pi}{2}$$ the solution of the given differential equation is $$y=\frac{4a}{\sqrt 3} \sin x -a$$
$$\frac{dy}{dx} \tan x =a+y , \quad y(\frac{\pi}{3})=a, \quad0\lt x \lt \frac{\pi}{2}$$ 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$$