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$\frac {dy}{dx} + 2y tan x = sin x $

$y (\frac {\pi}{3})=0$

${y_{\max }} = ?$

If y = y(x) is the solution to the differential equation, $\frac {dy}{dx} + 2y tan x = sin x $, $y (\frac {\pi}{3})=0$, then the maximum value of the function y(x) over R is equal to:

(A) $\frac {1}{2} $
(B) $\frac {1}{8} $
(C) $-\frac {15}{4}$
(D) 8

Solution

We have linear differential equation whose solution can be obtained from,

$y.IF = \int {IF.\sin xdx} $ where,

$IF = {e^{\int {2\tan xdx} }} = {e^{2\ln |\sec x|}} = {\sec ^2}x$

So, $y.{\sec ^2}x = \int {{{\sec }^2}x.\sin xdx}  = \int {\sec x.\tan xdx}  = \sec x + C$

$\because y\left( {\frac{\pi }{3}} \right) = 0,0.{\sec ^2}\frac{\pi }{3} = \sec \frac{\pi }{3} + C$

$\therefore C =  - 2$

So, $y.{\sec ^2}x = \sec x - 2$

$ \Rightarrow y = \frac{{\sec x - 2}}{{{{\sec }^2}x}} = \cos x - 2{\cos ^2}x$

$\therefore y =  - 2\left( {{{\cos }^2}x - \frac{1}{2}\cos x} \right) =  - 2\left[ {{{\left( {\cos x - \frac{1}{4}} \right)}^2} - \frac{1}{{16}}} \right]$

So, $y = \frac{1}{8} - (non - negative)$

$\therefore {y_{\max }} = \frac{1}{8}$

Answer: (B)