JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 14)
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:
$${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of the function y(x) over R is equal to:
$${1 \over 8}$$
8
$$-$$$${15 \over 4}$$
$${1 \over 2}$$
Explanation
$${{dy} \over {dx}} + 2\tan x.y = \sin x$$
$$I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x$$
$$y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} } $$
$$y{\sec ^2}x = \sec x + c$$
$$y = \cos x + c{\cos ^2}x$$
$$x = {\pi \over 3},y = 0$$
$$ \Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2$$
$$ \therefore $$ $$y = \cos x - 2{\cos ^2}x$$
$$y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)$$
$$y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$$
$$ \therefore $$ $${y_{\max }} = {1 \over 8}$$
$$I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x$$
$$y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} } $$
$$y{\sec ^2}x = \sec x + c$$
$$y = \cos x + c{\cos ^2}x$$
$$x = {\pi \over 3},y = 0$$
$$ \Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2$$
$$ \therefore $$ $$y = \cos x - 2{\cos ^2}x$$
$$y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)$$
$$y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$$
$$ \therefore $$ $${y_{\max }} = {1 \over 8}$$
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