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Solve a linear equation, then find a maximum or critical point of the solution

This type of question, asked 5 times: 2021 Paper 5, 25 Jun 2022, 26 Jun 2022, 26 Jul 2022, 23 Jan 2026 S2. All 5 questions 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

    25 Jun 2022

    75
    Let y=y(x)y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2(x+ 1)y^{'}-y=e^{3x}(x+ 1)^{2}, with y(0)=13y(0) =\frac{1}{3}. Then, the point x=−43x= -\frac{4}{3} for the curve y=y(x)y=y(x) is:

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  3. Question 3

    26 Jun 2022

    159
    If y=y(x)y = y(x) is the solution of the differential equation xdydx+2y=xx,y(1)=0x\frac{dy}{dx}+ 2y =x^{x},y(1) = 0 then the local maximum value of the function z(x)=x2y(x)−exz(x) =x^{2}y(x) -e^{x}, x∈Rx \in R is:

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  4. Question 4

    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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  5. Question 5

    23 Jan 2026 S2

    71
    If the solution curve y=f(x)y = f(x) of the differential equation (x2−4)y′−2xy+2x(4−x2)2=0, x>2\left( x^{2} - 4 \right)y' - 2xy + 2x\left( 4 - x^{2} \right)^{2} = 0,\ x > 2, passes through the point ( 3,15 ), then the local maximum value of f is ____\_\_\_\_