If y = y(x) is the solution of the differential equation, [dy dx + 2y x = x,y( 3 ) = 0], t…

Question

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

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

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

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

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

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

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

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

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

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

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

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

Solutions · 1