Question

Let y(x) be a solution of the initial value problem y'= y- y^2, y(0)= y0, 0 10 years ago

Answer 1

and also y has to be between 0 and 1..since
\[\frac{dy}{y-y^2}=dx \\ y \neq 0,1 \\ \text{ so we have three intervals } (-\infty,0) \text{ or } (0,1) \text{ or } (1,\infty) \\ \text{ but again since } 0<y_0<1 \text{ then } \\ 0<y<1 \\ \text{ so solve for } y \\ \text{ then show } y-y_0>0 \text{ for } x>0\]

Answer 2

I thought it was repetitive a bit so I deleted that one post :p

Answer 3

y= e^x/(e^x+c)