Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 5a + y, y(π/3) = 5a, 0 < x < π/2, where a is a constant.

[tex]y' \tan x = 5a + y \\ \\ \indent y' \tan x - 5a = y \\ \\ \indent \frac{y' \tan x}{\tan x} - \frac{5a}{\tan x} = \frac{y}{\tan x} \\ \\ \indent y' - 5a\cot x = y\cot x \\ \\ \indent \boxed{y' - y\cot x = 5a\cot x} \text{ (1)}[/tex]

Note that the above equation is a first order non-homogeneous linear differential equation which is of the form

[tex]y' + A(x)y = B(x)[/tex]

where (in the problem):

[tex]A(x) = -\cot x \\ B(x) = 5a\cot x[/tex]

We can solve this equation by multiplying both sides by the integrating factor I(x) which is calculated as 

[tex]I(x) = e^{\int {A(x)dx}} \\ \indent = e^{\int {-\cot x dx}} \\ \indent = e^{-\ln (\sin x) + C_0} \\ \indent = e^{(-1)(\ln (\sin x)) + C_0} \\ \indent = e^{C_0}\left (e^{\ln (\sin x)} \right )^{-1} \\ \indent = e^{C_0}\left (\sin x \right )^{-1} \\ \indent = e^{C_0}\left (\csc x \right ) \\ \indent \boxed{I(x) = K\csc x, K = e^{C_0}}[/tex]

Multiplying this integrating factor to both sides of equation (1), we have

[tex]y' - y\cot x = 5a\cot x \\ \indent K\csc x \left ( y' - y\cot x \right ) = K\csc x \left (5a\cot x \right ) \\ \indent \csc x \left ( y' - y\cot x \right ) = \csc x \left (5a\cot x \right ) \\ \indent \boxed{y'\csc x - y\csc x\cot x = 5a\csc x \cot x} \text{ (2)}[/tex]

Note that 

[tex]\boxed{y'\csc x - y\csc x\cot x = \frac{d}{dx}\left (y\csc x \right )}[/tex]

Using this information for equation (2), 

[tex]\frac{d}{dx}\left (y\csc x \right ) = 5a\csc x \cot x \\ \\ \indent y\csc x = \int {5a\csc x \cot x dx} \\ \\ \indent y\csc x = \int {5a \left ( \frac{1}{\sin x} \right )\left ( \frac{\cos x}{\sin x} \right ) dx} \\ \\ \indent \boxed{y\csc x = \int {5a \left ( \frac{\cos x}{\sin^2 x} \right ) dx}}\text{ (3)}[/tex]

Now, we evaluate the integral on the left side. Let [tex]u = \sin x, du = \cos x dx[/tex] so that 

[tex]\int {5a\left (\frac{\cos x}{\sin^2 x} \right ) dx} = \int {\frac{5a}{u^2}}du \\ \\ \indent = -\frac{5a}{u} + C \\ \\ \indent = -\frac{5a}{\sin x} + C \\ \\ \indent \boxed{\int {5a\left ( \frac{\cos x}{\sin^2 x} \right ) dx} = -5a\csc x + C} \text{ (4)}[/tex]

Using equation (4), equation (3) becomes

[tex]y\csc x = -5a\csc x + C \\ \\ \indent \frac{y\csc x}{\csc x} = \frac{-5a\csc x}{\csc x} + \frac{C}{\csc x} \\ \\ \indent \boxed{y = -5a + C\sin x} \text{ (5)} [/tex]

Since [tex]y(5\pi /3) = 5a[/tex], equation (5) becomes

[tex]y(5\pi /3) = -5a + C\sin (5\pi /3) \\ \\ 5a = -5a + C \left ( -\frac{\sqrt{3}}{2} \right ) \\ \\ C \left ( -\frac{\sqrt{3}}{2} \right ) = 10a \\ \\ \boxed{C = -\frac{20}{\sqrt{3}}}[/tex]

Using this value of the constant, therefore

[tex]\boxed{y = -5a - \frac{20}{\sqrt{3}}\sin x}[/tex]