How do you calculate the integral of xy' = y + √(x^2+y^2) as a homogeneous differential equation?

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  1. First, let v = y/x
    Now the equation is
    y' = v + √(1+v^2)
    Now, what about y'? That is dy/dx, but now we have a bunch of v stuff. Luckily,
    y = vx
    y' = xv' + v
    Now we have
    xv'+v = v + √(1+v^2)
    xv' = √(1+v^2)
    dv/√(1+v^2) = dx/x
    arcsinh(v) = lnx+c
    v = sinh(lnx+c)
    Now, v=y/x, so
    y = x sinh(lnx+c)
    That's pretty simple, but we can get rid of the sinh at the cost of a messier answer.
    Now, recall that sinh(z) = (e^z - e^-z)/2 so
    y = x (e^(lnx+c) - e^-(lnx+c))/2
    Now, since e^lnx = x, e^(lnx+c) = xe^c
    y = x(xe^c - 1/(xe^c))/2
    = x(x^2e^2c-1)/(2xe^c)
    = ((xe^c)^2-1)/(2e^c)
    I think I prefer the sinh form.

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    posted by oobleck

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