MATH 4A Lecture Notes - Lecture 3: Constant Function, Partial Derivative
Document Summary
Not every ivp has a solution at all, and some that do have solutions have more than one. y"=x/(-y); y(3)=0 doesn"t have unique solution. x/(-y) is discontinuous across y=0. This jump in values for the rhs function means there doesn"t exist a solution that crosses the line y=0. Four conditions for existing a unique solution to the ivp. 1. f is continuous in t on ( , ) 2. f is continuous in y on ( , ). 4. df/dy is continuous in y on ( , ). y"=sqrt(y) y"/sqrt(y)dy=1dt. We don not have uniqueness here. y(t)=0 is a solution. y(t)=0 for 0=
