QJEEM21Mae8c5: JEE Main 2021 Mathematics - Class 12 JEEM21Mae8c5 Solution: If y = y(x) is the solution of the differential equation, ,…

Question

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If y = y(x) is the solution of the differential equation,

{{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0

, then the maximum value of the function y(x) over R is equal to:

Accepted Answer

{{dy} \over {dx}} + 2\tan x.y = \sin x

I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x

y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} }

y{\sec ^2}x = \sec x + c

y = \cos x + c{\cos ^2}x

x = {\pi \over 3},y = 0

\Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2

\therefore

y = \cos x - 2{\cos ^2}x

y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)

y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}

\therefore

{y_{\max }} = {1 \over 8}

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