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Solve for particular solution. dy/dx + 2y tan x = sin x; y = 0 when x = pi/3

Solve for particular solution.

    Q13.    \frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}

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Given equation is
\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}
This is  \frac{dy}{dx} + py = Q  type where p = 2\tan x and Q = \sin x
Now,
I.F. = e^{\int pdx}= e^{\int 2\tan xdx}= e^{2\log |\sec x|}= \sec^2 x                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sec^2 x) =\int ((\sin x)\times \sec^2 x)dx +C
y(\sec^2 x) =\int (\sin \times \frac{1}{\cos x}\times \sec x)dx +C\\ \\ y(\sec^2 x) = \int \tan x\sec xdx+ C\\ \\ y.\sec^2 x= \sec x+C
Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when x= \frac{\pi}{3}
at  x= \frac{\pi}{3}
0.\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C\\ \\ C = - 2
Now,

   y.\sec^2 x= \sec x - 2\\ \frac{y}{\cos ^2x}= \frac{1}{\cos x}- 2\\ y = \cos x- 2\cos ^2 x
Therefore, the particular solution is y = \cos x- 2\cos ^2 x

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