How Does the Initial Condition y(0)=1 Affect the Solution to dy/dx=(y-1)^2?

[bcjochim07]

dy/dx = (y-1)^2 given the initial condition y(0) =1 find an explicit solution.


separation of variables gives:

(y-1)^-2 dy = dx

integrating both sides I get

-(y-1)^-1 = x + c

-1= (y-1)(x+c)

-1/(x+c) = y - 1

y= 1 - 1/(x+c) but how can the intial condition work?

no value of c will make 1= 1 - 1(0+c)

 

[NoMoreExams]

Pretty sure y = 1 is a fixed point (or a collection thereof, the whole y = 1 line) so you can't have any solution involving that condition since all solutions are basically just integral curves. Therefore all solutions/integral curves will either approach the y = 1 line or go away from it depending on its stability.