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x^2 dy + (y - 1/x) dx = 0 (or with x and y swapped): find the value at a point

The same question, asked 3 times: 2021 Paper 24, 24 Jan 2023, 22 Jan 2025. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 24

    67
    If y=y(x)y = y(x) is the solution curve of the differential equation x2dy+(y−1x)dx=0 ;x>0x^{2}dy +\left( y -\frac{1}{x} \right)dx = 0\ ;x > 0 and y(1)=1y(1) = 1, then y(12)y\left( \frac{1}{2} \right) is equal to:

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  2. Question 2

    24 Jan 2023

    70
    Let y=y(x)y = y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(12)=3−ex^{3}dy + (xy - 1)dx = 0,x > 0,y\left( \frac{1}{2} \right)= 3 - e. Then y(1)y(1) is equal to

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  3. Question 3

    22 Jan 2025

    62
    Let x=x(y)x = x(y) be the solution of the differential equation y2dx+(x−1y)dy=0y^{2}dx +\left( x -\frac{1}{y} \right)dy = 0. If x(1)=1x(1) = 1, then x(12)x\left( \frac{1}{2} \right) is :

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