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    I need to find the general solution of the differential equation: \frac{dy}{dx} = 1+y^2

    My workings:

    \frac{dx}{dy} = \frac{1}{1+y^2}

    (can't get the next bit to work in LaTeX so sorry about this)

    x = integral of 1/(1+y^2) dy
    = inverse tan(y) + k

    x-k = inverse tan (y)

    y = tan(x-k)

    but the answer should be y = tan(x + Pi/4 -2)

    Any ideas where I'm going wrong? Thanks

    Your answer isn't wrong.

    y = tan(x-k) is the general solution of the differential equation.

    y = tan(x + Pi/4 -2) is the general solution subject to a condition i.e. when x = ... , y = ...
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