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:
Solution
${{dy} \over {dx}} + 2\tan x.y = \sin x$<br><br>$I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x$<br><br>$y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} }$<br><br>$y{\sec ^2}x = \sec x + c$<br><br>$y = \cos x + c{\cos ^2}x$<br><br>$x = {\pi \over 3},y = 0$<br><br>$\Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2$<br><br>$\therefore$ $y = \cos x - 2{\cos ^2}x$<br><br>$$y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)$$<br><br>$y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$<br><br>$\therefore$ ${y_{\max }} = {1 \over 8}$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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