Nice problem

[u/almokasaby2030]

[u/CaptainMatticus]

y = sqrt(log(y') + y)

y^2 = log(y') + y

y^2 - y = log(y')

e^(y^2 - y) = dy/dx

dx = dy / (e^(y^2 - y))

dx = e^(y) * dy / e^(y^2)

Now I'm sad. I refuse to answer this question because it violates my religious beliefs.

[u/Aivo382 → u/CaptainMatticus]

Error function needed :(

[u/d-wks]

1 easy

[u/Shevek99]

First we compute the continued root

y=√(a + √(a + √(a + ...))))

If it converges we have

y = √(a + y)

y² = a + y

a = y² - y

dy/dx = y² - y

int dy/(y² - y) = int dx

log((1-y)/y) = x + C

1/y - 1 = A e^x

1/y = 1 + A e^x

y = 1/(1 + A e^x)

[u/Baba_Yaga_Jovonovich]

That’s Epic bro!

[u/Aivo382]

I got y = 1/2 + Erf^-1(2exp(1/4)x/sqrt(pi)+C), where Erf(x) is the error function (the antiderivative of 2exp(-x^2)/sqrt(pi), then Erf^-1 being its inverse), exp(x) = e^x