This differential equation is separable.
dy/(dx) = 10-2y
dy = (10-2y) dx
1/(10-2y) dy = dx
int dy/(10-2y) = int dx
Let u = 10-2y -> du = -2 dy -> -1/2 du = dy
-1/2 int (du)/(u) = int x dx
-1/2 ln(10-2y) = x + C
ln(10-2y) = -2(x+C)
10-2y = e^(-2(x+C))
-2y+10 = e^(-2(x+C))
-2y = e^(-2(x+C)) - 10
color(red)(-2y = e^(-2x)e^(-2C) - 10)
color(red)(-2y = C e^(-2x) - 10)
color(red)(y = C/(-2) e^(-2x) - 10/(-2))
color(red)(y = C e^(-2x) +5)
y = - (e^(-2(x+C)) -10)/(2)
y = (e^(-2(x+C)) + 10)/(2)