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Separable equation with powers 2^x and 2^y

This type of question, asked 3 times: 2021 Paper 22, 2021 Paper 23, 27 Jun 2022. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 22

    76
    If dydx=2x+y−2x2y,y(0)=1\frac{dy}{dx}=\frac{2^{x+y}-2^{x}}{2^{y}},y(0) = 1, then y(1)y(1) is equal to :

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  2. Question 2

    2021 Paper 23

    70
    If dydx=2xy+2y⋅2x2x+2x+ylog⁡e2,y(0)=0\frac{dy}{dx}=\frac{2^{x}y+2^{y}\cdot2^{x}}{2^{x}+2^{x+y}\log_{e}2},y(0) = 0, then for y=1y= 1, the value of xx lies in the interval:

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  3. Question 3

    27 Jun 2022

    71
    If dydx+2x−y(2y−1)2x−1=0,x,y>0,y(1)=1\frac{dy}{dx}+\frac{2^{x - y}\left( 2^{y}- 1 \right)}{2^{x}- 1}= 0,x,y > 0,y(1) = 1, then y(2)y(2) is equal to :

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