(21 pts) Solve the IVP:

xy' = (1 - y

^{2})^{1/2}, y(1) = 0.### Solution

This is a separable equation: dy/(1 - y

^{2})^{1/2}= dx/x. Integrating both sides be get:arcsin(y) = ln|x| + C. With the usual trick of replacing an arbitrary constant by another one, this reduces to:

e ^{arcsin(y)}= Cx,so that we found x as a function of y. That's fine. With differential equations you often do not know how the solution will come out. The initial conditions gives: e

^{0}= C, i.e. C = 1.### Answer: e

^{arcsin(y)}= x.### P.S.

Please note. This problem has inherent ambiguity due to my carelessness. The initial condition falls on the boundary of the domain of definition of the right hand side of the equation. I should have notice that - I did not. For this reason I shall accept as correct any solution that reaches

and shall disregard any treatment ofarcsin(y) = ln|x| + C *arcsin*that comes afterwards.Note also that getting an implicit form of a solution is very acceptable. There's no requirement to present "

*y as a function of x*." I have mentioned that several times in class.