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x dy/dx - y = sqrt(y^2 + a x^2) with y(1) given: find y at another point

The same question, asked 3 times: 29 Jun 2022*, 28 Jul 2022*, 22 Jan 2026 S1. Every wording below, oldest first.

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  1. Question 1

    29 Jun 2022

    62
    Let the solution curve of the differential equation xdydx−y=y2+16x2,y(1)=3x\frac{dy}{dx}- y =\sqrt{y^{2}+ 16x^{2}},y(1) = 3 be y=y(x)y = y(x). Then y(2)y(2) is equal to:

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  2. Question 2

    28 Jul 2022

    61
    Let the solution curve of the differential equation xdy=(x2+y2+y)dx,x>0,xdy =\left( \sqrt{x^{2}+y^{2}}+ y \right)dx,x > 0, intersect the line x=1x = 1 at y=0y = 0 and the line x=2x = 2 at y=αy = \alpha. Then the value of α\alpha is:

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  3. Question 3

    22 Jan 2026 S1

    69
    Let the solution curve of the differential equation xdy−ydx=x2+y2dx,x>0,y(1)=0xdy - ydx =\sqrt{x^{2}+y^{2}}dx,x > 0,y(1) = 0, be y=y(x)y = y(x). Then y(3)y(3) is equal to

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