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dy/dx + 2y tan x = sin x with y(pi/3) = 0: find the maximum of y or a value

The same question, asked 3 times: 2021 Paper 5, 26 Jul 2022, 24 Jan 2025. Every wording below, oldest first.

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  1. Question 1

    2021 Paper 5

    77
    If y=y(x)y=y(x) is the solution of the differential equation, dydx+2ytanx=sin⁡x,y(π3)=0\frac{dy}{dx}+ 2ytanx= \sin x,y\left( \frac{\pi}{3} \right)= 0, then the maximum value of the function y(x)y(x) over R is equal to :

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  2. Question 2

    26 Jul 2022

    71
    If dydx+2ytanx=sin⁡x,0<x<π2\frac{dy}{dx}+ 2ytanx = \sin x,0 < x <\frac{\pi}{2} and y(π3)=y\left( \frac{\pi}{3} \right)=0, then the maximum value of y(x)y(x) is

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  3. Question 3

    24 Jan 2025

    148
    Let y=y(x)y=y(x) be the solution of the differential equation 2cos⁡xdydx=sin⁡2x−4ysinx,x∈(0,π2)2\cos x\frac{dy}{dx}= \sin2x- 4ysinx,x\in\left( 0,\frac{\pi}{2} \right). If y(π3)=0y\left( \frac{\pi}{3} \right)= 0, then y′(π4)+y(π4)y^{'}\left( \frac{\pi}{4} \right)+y\left( \frac{\pi}{4} \right) is equal to _____\_\_\_\_\_ .