Solving IVP: \[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0

[ssb]

Homework Statement

Please take a look at my work and help me figure out where I went wrong. Thanks!
Use separation of variables to solve the IVP
\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0

Homework Equations

The Attempt at a Solution

Use separation of variables to solve the IVP
\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0

dy/dt = \sqrt{1-y^2}/t

1/\sqrt{1-y^2} dy = 1/t dt

\int 1/\sqrt{1-y^2} dy = \int 1/t dt

arcsin(y) = ln (abs (t)) + C

C = 1

therefore the formula should be

y = sin(ln(abs(t))+1)

Thanks

 

[Gib Z]

I don't understand what you think is the problem.

 

[d_leet]

I don't see a problem except that your value of C is incorrect, because with that choice y(1) is not equal to 1.

 

[HallsofIvy]

arcsin(1) is \pi/2, not 1.

I will also point out that this is of the form y'= \sqrt{1- y^2}/t and the function is NOT Lipshchitz in y on any interval containing y= 1. Therefore, the solution is not unique. y= 1 for all t is another obvious solution and, in fact, there are an infinite number of solutions.