Need hep with Initial Value Problem

Question

Hello I keep getting stuck on these two problems and can't seem to figure out how to solve it

1. y'=2-y , y(0)=1
2. y'=y(2-y) , y(0)=1

Answer 1

dy/dx = 2 - y

(1/(2-y))dy = dx

∫(1 / (2-y))dy = ∫dx

-ln l2-yl = x + K

ln l 2-y l = -x - K

l 2 - y l = e^{-x- K} = e^-Ke^-x

2 - y = ±e^-Ke^-x = Ce^-x

y = 2 - Ce^-x

When x = 0, y = 1. So, 1 = 2 - C. C = 1

Therefore, y = 2 - e^-x

------------------------------------

dy/dx = y(2 - y)

1 / [y(2-y)] dy = dx

Integrate by Partial Fractions to get (1/2)lnlyl - (1/2)lnl2-yl = x + K

ln l y / (2-y )l = 2x + 2K

l y / (2-y) l = e^2x+2K = e^2Ke^2x

y / (2-y) = ±e^2ke^2x = Ce^2x

When x = 0, y = 1. So, C = 1.

y / (2 - y) = e^2x

y = 2e^2x - ye^2x

y + ye^2x = 2e^2x

y(1 + e^2x) = 2e^2x

So, y = 2e^2x / (1 + e^2x)