Let y=y (x ) be the solution of the differential equation 2 cosx dy dx= sin2x-4y sinx ,x (0 , 2 ). If y ( 3 )=0, then y^ ' ( 4 )+y ( 4 ) is equal to .

Question

Accepted Answer

Correct answer: 1. (1) dy dx+2y tan ⁡x= sin ⁡x I .F .=e^2 tan ⁡ xdx= sec^2 ⁡x y sec^2 ⁡x= sin ⁡x cos^2 ⁡x dx = tanx sec ⁡ xdx = sec ⁡x+CC=-2 y= cosx-2 cos^2x y ( 4 )= 1 2-1 y^ '=- sin ⁡x+4 cos ⁡x sin ⁡x y^ ' ( 4 )=- 1 2+2 y^ ' ( 4 )+y ( 4 )=1

$Let y = y (x) be the solution of the differential equation 2 co s x \frac{d y}{d x} = si n 2 x - 4 y sin x, x (0, \frac{π}{2}) . If y (\frac{π}{3}) = 0, then y^{’} (\frac{π}{4}) + y (\frac{π}{4}) is equal to .$

Solution

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