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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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