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Solve the differential and initial value problem:

x(dy/dx) + 1 = y^2


I tried using bernoulli and it didn't quite work.

  • Calculus -

    xy' + 1 = y^2
    xy' = y^2-1
    y' = (y^2-1)/x
    dy/(y^2-1) = dx/x
    arctanh(y) = log(x)
    log (1-y)/(1+y) = 2log(x)+c

    and you can massage that into exponentials and wind up with

    y = 1-e^(2cx^2) / 1+e^(2cx^2)

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