A differential equation problem

Shad
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I'm not sure how to solve the following problem. It's quite unlike the other problems I've had to solve since I started learning about differential equations. Can anybody help?

A disease spreads at a speed that is proportional to the multiplier of healthy and sick people. Let's call the sick quotient of a group x and the quotient of healthy people 1 - x. Then we have dx / dt = kx(1 - x). If half of the group is sick and the disease spreads at a constant speed, every member of the group will be sick in 24 days time. Then we have x = 1/2 , 1 - x = 1/2 and dx / dt = (1/2)/24 = 1/48 , so 1/48 = k * 1/2 * 1/2 <=> k = 1/12 , which gives dx / dt = 1/12 x(1 - x). Calculate how big a fraction of the group is sick in 12 days time.
 
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Hint: Separate the variables and then integrate, do you know how to integrate partial fractions?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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